#
Multiple Choice Questions
One and More than One Correct
1. If the bond length of CO bond in carbon monoxide is 1.128 Ã…, then what is the value of CO bond length in $\text{Fe(CO)}_5$? [JEE 2006]
a. 1.15 Ã…
b. 1.128 Ã…
c. 1.72 Ã…
d. 1.118 Ã…
Answer: (a)
Solution:
In $\text{Fe(CO)}_5$, due to synergic bonding ($\text{d}\pi - \text{p}\pi$ back-bonding) from the central metal ion into the antibonding $\pi^*$ orbital of $\text{CO}$, the $\text{C-O}$ bond order decreases, resulting in an increase in the $\text{C-O}$ bond length compared to free $\text{CO}$ ($1.128\text{ \AA}$). Thus, the bond length becomes $1.15\text{ \AA}$.
2. Among the following metal carbonyls, the C----O bond order is lowest in [JEE 2007 P-II]
a. $[\text{Mn(CO)}_6]^+$
b. $[\text{Fe(CO)}_5]$
c. $[\text{Cr(CO)}_6]$
d. $[\text{V(CO)}_6]^-$
Answer: (d)
Solution:
Greater the negative charge on the central metal atom, higher is the extent of back-bonding ($\text{d}\pi - \text{p}\pi$ donation to $\text{CO}$). This weakens the $\text{C-O}$ bond and reduces its bond order. Among all the given species, $[\text{V(CO)}_6]^-$ has the highest negative charge on the central metal, leading to maximum back-bonding and the lowest $\text{C-O}$ bond order.
3. Both $\text{Ni(CO)}_4$ and $[\text{Ni(CN)}_4]^{2-}$ are diamagnetic. The hybridization of nickel in these complexes, respectively, are [JEE 2008 P-II]
a. $\text{sp}^3, \text{sp}^3$
b. $\text{sp}^3, \text{dsp}^2$
c. $\text{dsp}^2, \text{sp}^3$
d. $\text{dsp}^2, \text{dsp}^2$
Answer: (b)
Solution:
In $\text{Ni(CO)}_4$, $\text{Ni}$ is in $0$ oxidation state ($3\text{d}^8 4\text{s}^2$). Strong field ligand $\text{CO}$ pairs up electrons into $3\text{d}^{10}$, making hybridization $\text{sp}^3$ (tetrahedral).
In $[\text{Ni(CN)}_4]^{2-}$, $\text{Ni}^{2+}$ is in $+2$ oxidation state ($3\text{d}^8$). Strong field ligand $\text{CN}^-$ forces pairing to form a square planar complex with $\text{dsp}^2$ hybridization.
4. The spin only magnetic moment value (in Bohr magneton units) of $\text{Cr(CO)}_6$ is [JEE 2009 P-II]
a. 0
b. 2.84
c. 4.90
d. 5.92
Answer: (a)
Solution:
In $\text{Cr(CO)}_6$, Chromium is in $0$ oxidation state ($3\text{d}^5 4\text{s}^1$). The strong field ligand $\text{CO}$ causes complete pairing of all 6 electrons into $3\text{d}$ subshell ($t_{2g}^6 e_g^0$). Thus, the number of unpaired electrons ($n$) is 0, so $\mu_{\text{spin-only}} = 0\text{ B.M.}$
5. The correct structure of ethylenediaminetetraacetic acid (EDTA) is [JEE 2010 P-I]
a.
b.
c.
d.
Answer: (c)
Solution:
Ethylenediaminetetraacetic acid (EDTA) contains two amino nitrogens and four carboxylic acid groups attached via ethylene bridge: $(\text{HOOC-CH}_2)_2\text{N-CH}_2-\text{CH}_2-\text{N}(\text{CH}_2-\text{COOH})_2$.
6. The ionization isomer of $[\text{Cr}(\text{H}_2\text{O})_4\text{Cl}(\text{NO}_2)]\text{Cl}$ is [JEE 2010 P-I]
a. $[\text{Cr}(\text{H}_2\text{O})_4(\text{O}_2\text{N})]\text{Cl}_2$
b. $[\text{Cr}(\text{H}_2\text{O})_4\text{Cl}_2](\text{NO}_2)$
c. $[\text{Cr}(\text{H}_2\text{O})_4\text{Cl}(\text{ONO})]\text{Cl}$
d. $[\text{Cr}(\text{H}_2\text{O})_4\text{Cl}_2(\text{NO}_2)]\text{H}_2\text{O}$
Answer: (b)
Solution:
Ionization isomers exchange ions inside and outside the coordination sphere. $[\text{Cr}(\text{H}_2\text{O})_4\text{Cl}_2](\text{NO}_2)$ exchanges $\text{Cl}^-$ with $\text{NO}_2^-$.
7. The complex showing a spin-only magnetic moment of 2.82 B.M. is [JEE 2010 P-II]
a. $\text{Ni(CO)}_4$
b. $[\text{NiCl}_4]^{2-}$
c. $\text{Ni(PPh}_3)_4$
d. $[\text{Ni(CN)}_4]^{2-}$
Answer: (b)
Solution:
$$\mu = \sqrt{n(n+2)} = 2.82\text{ B.M.} \implies n = 2$$In $[\text{NiCl}_4]^{2-}$, $\text{Ni}^{2+}$ ($3\text{d}^8$) with weak field ligand $\text{Cl}^-$ has tetrahedral geometry with 2 unpaired electrons.
8. Geometrical shapes of the complexes formed by the reaction of $\text{Ni}^{2+}$ with $\text{Cl}^-$, $\text{CN}^-$ and $\text{H}_2\text{O}$ respectively, are [JEE 2011 P-I]
a. octahedral, tetrahedral and square planar
b. tetrahedral, square planar and octahedral
c. square planar, tetrahedral and octahedral
d. octahedral, square planar and octahedral
Answer: (b)
Solution:
1. $[\text{NiCl}_4]^{2-}$: $\text{sp}^3$, Tetrahedral
2. $[\text{Ni(CN)}_4]^{2-}$: $\text{dsp}^2$, Square planar
3. $[\text{Ni}(\text{H}_2\text{O})_6]^{2+}$: $\text{sp}^3\text{d}^2$, Octahedral
9. Among the following complexes (K–P):
$\text{K}_3[\text{Fe(CN)}_6]$ (K), $\left[\text{Co}(\text{NH}_3)_6\right]\text{Cl}_3$ (L), $\text{Na}_3[\text{Co(oxalate)}_3]$ (M), $[\text{Ni}(\text{H}_2\text{O})_6]\text{Cl}_2$ (N), $\text{K}_2[\text{Pt(CN)}_4]$ (O) and $[\text{Zn}(\text{H}_2\text{O})_6](\text{NO}_3)_2$ (P)
Following compounds are diamagnetic. [JEE 2011 P-II]
a. K, L, M, N
b. K, M, O, P
c. L, M, O, P
d. L, M, N, O
Answer: (c)
Solution:
$\text{L: } [\text{Co}(\text{NH}_3)_6]^{3+} \implies \text{Co}^{3+} (3\text{d}^6)$, strong field ligand $\implies$ diamagnetic ($n=0$)
$\text{M: } [\text{Co}(\text{ox})_3]^{3-} \implies \text{Co}^{3+} (3\text{d}^6)$, strong field ligand $\implies$ diamagnetic ($n=0$)
$\text{O: } [\text{Pt(CN)}_4]^{2-} \implies \text{Pt}^{2+} (5\text{d}^8)$, square planar $\implies$ diamagnetic ($n=0$)
$\text{P: } [\text{Zn}(\text{H}_2\text{O})_6]^{2+} \implies \text{Zn}^{2+} (3\text{d}^{10}) \implies$ diamagnetic ($n=0$)
10. $\text{NiCl}_2\left\{\text{P}(\text{C}_2\text{H}_5)_2(\text{C}_6\text{H}_5)\right\}_2$ exhibits temperature dependent magnetic behavior (paramagnetic / diamagnetic). The coordination geometries of $\text{Ni}^{2+}$ in the paramagnetic and diamagnetic states are respectively. [JEE 2012 P-II]
a. tetrahedral and tetrahedral
b. square planar and square planar
c. tetrahedral and square planar
d. square planar and tetrahedral
Answer: (c)
Solution:
For $\text{Ni}^{2+}$ ($3\text{d}^8$):
Paramagnetic state corresponds to 2 unpaired electrons in tetrahedral geometry ($\text{sp}^3$).
Diamagnetic state corresponds to 0 unpaired electrons in square planar geometry ($\text{dsp}^2$).
11. Consider the following complex ions, P, Q and R: $\text{P} = [\text{FeF}_6]^{3-}$, $\text{Q} = [\text{V}(\text{H}_2\text{O})_6]^{2+}$ and $\text{R} = [\text{Fe}(\text{H}_2\text{O})_6]^{2+}$. The correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.) is [JEE Adv. 2013 P-I]
a. R < Q < P
b. Q < R < P
c. R < P < Q
d. Q < P > R
Answer: (b)
Solution:
$\text{P: } [\text{FeF}_6]^{3-} \implies \text{Fe}^{3+} (3\text{d}^5)$, weak field ligand $\implies n = 5 \implies \mu = 5.92\text{ B.M.}$
$\text{Q: } [\text{V}(\text{H}_2\text{O})_6]^{2+} \implies \text{V}^{2+} (3\text{d}^3) \implies n = 3 \implies \mu = 3.87\text{ B.M.}$
$\text{R: } [\text{Fe}(\text{H}_2\text{O})_6]^{2+} \implies \text{Fe}^{2+} (3\text{d}^6)$, weak field ligand $\implies n = 4 \implies \mu = 4.90\text{ B.M.}$
Order: $\text{Q} < \text{R} < \text{P}$
12. Which of the following complex species is not expected to exhibit optical isomerism? [JEE Main 2013]
a. $[\text{Co(en)}_3]^{3+}$
b. $[\text{Co(en)}_2\text{Cl}_2]^+$
c. $[\text{Co}(\text{NH}_3)_3\text{Cl}_3]$
d. $[\text{Co(en)}(\text{NH}_3)_2\text{Cl}_2]^+$
Answer: (c)
Solution:
$[\text{Co}(\text{NH}_3)_3\text{Cl}_3]$ is of the type $[\text{MA}_3\text{B}_3]$ which exists as facial (fac) and meridional (mer) isomers. Both possess a plane of symmetry ($\text{POS}$) and are optically inactive.
13. The number of geometric isomers that can exist for square planar $[\text{Pt}(\text{Cl})(\text{py})(\text{NH}_3)(\text{NH}_2\text{OH})]^\oplus$ is (py = pyridine): [JEE Main 2015]
a. 2
b. 3
c. 4
d. 6
Answer: (b)
Solution:
Square planar complex of type $[\text{Mabcd}]$ forms 3 geometrical isomers.
14. The pair having the same magnetic moment is: [At. No.: Cr = 24, Mn = 25, Fe = 26, Co = 27] [JEE Main 2016]
a. $[\text{Cr}(\text{H}_2\text{O})_6]^{2+}$ and $[\text{Fe}(\text{H}_2\text{O})_6]^{2+}$
b. $[\text{Mn}(\text{H}_2\text{O})_6]^{2+}$ and $[\text{Cr}(\text{H}_2\text{O})_6]^{2+}$
c. $[\text{CoCl}_4]^{2-}$ and $[\text{Fe}(\text{H}_2\text{O})_6]^{2+}$
d. $[\text{Cr}(\text{H}_2\text{O})_6]^{2+}$ and $[\text{CoCl}_4]^{2-}$
Answer: (a)
Solution:
$[\text{Cr}(\text{H}_2\text{O})_6]^{2+} \implies \text{Cr}^{2+} (3\text{d}^4)$, high spin $\implies n = 4$ unpaired electrons.
$[\text{Fe}(\text{H}_2\text{O})_6]^{2+} \implies \text{Fe}^{2+} (3\text{d}^6)$, high spin $\implies n = 4$ unpaired electrons.
Since both have $n = 4$ unpaired electrons, they have the same spin-only magnetic moment ($\mu = \sqrt{24}\text{ B.M.} = 4.90\text{ B.M.}$).
15. Which one of the following complexes shows optical isomerism? [JEE Main 2016]
a. $[\text{Co}(\text{NH}_3)_3\text{Cl}_3]$
b. $\text{cis}-[\text{Co(en)}_2\text{Cl}_2]\text{Cl}$
c. $\text{trans}-[\text{Co(en)}_2\text{Cl}_2]\text{Cl}$
d. $[\text{Co}(\text{NH}_3)_4\text{Cl}_2]\text{Cl}$
Answer: (b)
Solution:
$\text{cis}-[\text{Co(en)}_2\text{Cl}_2]^+$ lacks a plane of symmetry ($\text{POS}$) and center of inversion ($\text{COS}$), making it non-superimposable on its mirror image, thus showing optical isomerism.
16. Among $\text{Ni(CO)}_4$, $[\text{NiCl}_4]^{2-}$, $[\text{Co}(\text{NH}_3)_4\text{Cl}_2]\text{Cl}$, $\text{Na}_3[\text{CoF}_6]$, $\text{Na}_2\text{O}_2$ and $\text{CsO}_2$, the total number of paramagnetic compounds is [JEE Adv. 2016 P-I]
a. 2
b. 3
c. 4
d. 5
Answer: (c)
Solution:
1. $\text{Ni(CO)}_4$: $\text{Ni}^0 (3\text{d}^{10}) \implies$ Diamagnetic
2. $[\text{NiCl}_4]^{2-}$: $\text{Ni}^{2+} (3\text{d}^8) \implies n = 2 \implies$ Paramagnetic
3. $[\text{Co}(\text{NH}_3)_4\text{Cl}_2]\text{Cl}$: $\text{Co}^{3+} (3\text{d}^6)$, low spin $\implies$ Diamagnetic
4. $\text{Na}_3[\text{CoF}_6]$: $\text{Co}^{3+} (3\text{d}^6)$, high spin ($\text{F}^-$ is weak) $\implies n = 4 \implies$ Paramagnetic
5. $\text{Na}_2\text{O}_2$: $\text{O}_2^{2-}$ peroxide ion $\implies$ Diamagnetic
6. $\text{CsO}_2$: $\text{O}_2^-$ superoxide ion $\implies 1$ unpaired electron $\implies$ Paramagnetic
Total paramagnetic compounds = 3 ($[\text{NiCl}_4]^{2-}$, $\text{Na}_3[\text{CoF}_6]$, $\text{CsO}_2$).
17. The geometries of the ammonia complexes of $\text{Ni}^{2+}$, $\text{Pt}^{2+}$ & $\text{Zn}^{2+}$, respectively, are [JEE Adv. 2016 P-II]
a. octahedral, square planar and tetrahedral
b. square planar, octahedral and tetrahedral
c. tetrahedral, square planar and octahedral
d. octahedral, tetrahedral and square planar
Answer: (a)
Solution:
1. $[\text{Ni}(\text{NH}_3)_6]^{2+}$: Octahedral ($\text{sp}^3\text{d}^2$)
2. $[\text{Pt}(\text{NH}_3)_4]^{2+}$: Square planar ($\text{dsp}^2$, $5\text{d}$ metal)
3. $[\text{Zn}(\text{NH}_3)_4]^{2+}$: Tetrahedral ($\text{sp}^3$, $3\text{d}^{10}$)
18. The compound(s) that exhibit(s) geometrical isomerism is(are) [JEE 2009 P-I]
a. $[\text{Pt(en)}\text{Cl}_2]$
b. $[\text{Pt(en)}_2]\text{Cl}_2$
c. $[\text{Pt(en)}_2\text{Cl}_2]\text{Cl}_2$
d. $[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]$
Answer: (c, d)
Solution:
c. $[\text{Pt(en)}_2\text{Cl}_2]^{2+}$ is octahedral and exists in cis and trans forms.
d. $[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]$ is square planar and exists in cis and trans forms.
19. The pair(s) of coordination complexes/ions exhibiting the same kind of isomerism is(are) [JEE Adv. 2013 P-I]
a. $[\text{Cr}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2$ and $[\text{Cr}(\text{NH}_3)_4\text{Cl}_2]\text{Cl}$
b. $[\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+$ and $[\text{Pt}(\text{NH}_3)_2(\text{H}_2\text{O})\text{Cl}]^+$
c. $[\text{CoBr}_2\text{Cl}_2]^{2-}$ and $[\text{PtBr}_2\text{Cl}_2]^{2-}$
d. $[\text{Pt}(\text{NH}_3)_3(\text{NO}_2)]\text{Cl}$ and $[\text{Pt}(\text{NH}_3)_3\text{Cl}]\text{Br}$
Answer: (b, d)
Solution:
b. Both $[\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+$ and $[\text{Pt}(\text{NH}_3)_2(\text{H}_2\text{O})\text{Cl}]^+$ exhibit geometrical isomerism (cis-trans).
d. Both $[\text{Pt}(\text{NH}_3)_3(\text{NO}_2)]\text{Cl}$ and $[\text{Pt}(\text{NH}_3)_3\text{Cl}]\text{Br}$ exhibit ionization isomerism.
20. On treatment of 100 mL of 0.1 M solution of $\text{CoCl}_3 \cdot 6\text{H}_2\text{O}$ with excess $\text{AgNO}_3$, $1.2 \times 10^{22}$ ions are precipitated. The complex is [JEE Main 2017]
a. $[\text{Co}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2 \cdot \text{H}_2\text{O}$
b. $[\text{Co}(\text{H}_2\text{O})_4\text{Cl}_2]\text{Cl} \cdot 2\text{H}_2\text{O}$
c. $[\text{Co}(\text{H}_2\text{O})_3\text{Cl}_3] \cdot 3\text{H}_2\text{O}$
d. $[\text{Co}(\text{H}_2\text{O})_6]\text{Cl}_3$
Answer: (a)
Solution:
$$\text{Moles of complex} = 0.1 \times 0.1 = 0.01\text{ mol}$$ $$\text{Moles of } \text{AgCl precipitated} = \frac{1.2 \times 10^{22}}{6.02 \times 10^{23}} \approx 0.02\text{ mol}$$ $$\text{Number of ionizable } \text{Cl}^- \text{ ions per complex molecule} = \frac{0.02}{0.01} = 2$$Therefore, the formula of the complex is $[\text{Co}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2 \cdot \text{H}_2\text{O}$.
21. Addition of excess aqueous ammonia to a pink coloured aqueous solution of $\text{MCl}_2 \cdot 6\text{H}_2\text{O}$ (X) and $\text{NH}_4\text{Cl}$ gives an octahedral complex Y in the presence of air. In aqueous solution, complex Y behaves as 1 : 3 electrolyte. The reaction of X with excess HCl at room temperature results in the formation of a blue coloured complex Z. The calculated spin only magnetic moment of X and Z is 3.87 B.M., whereas it is zero for complex Y. Among the following options, which statement(s) is(are) correct? [JEE Adv. 2017 P-I]
a. The hybridization of the central metal ion in Y is $\text{d}^2\text{sp}^3$
b. When X and Z are in equilibrium at 0°C, the colour of the solution is pink
c. Z is a tetrahedral complex
d. Addition of silver nitrate to Y gives only two equivalents of silver chloride
Answer: (a, b, c)
Solution:
$\text{X} = [\text{Co}(\text{H}_2\text{O})_6]\text{Cl}_2$ (pink, $\text{Co}^{2+} 3\text{d}^7$, $n=3$, $\mu = 3.87\text{ B.M.}$)
$\text{Y} = [\text{Co}(\text{NH}_3)_6]\text{Cl}_3$ (1:3 electrolyte, $\text{Co}^{3+} 3\text{d}^6$, low spin diamagnetic, $\text{d}^2\text{sp}^3$)
$\text{Z} = [\text{CoCl}_4]^{2-}$ (blue, tetrahedral, $\text{Co}^{2+} 3\text{d}^7$, $n=3$, $\mu = 3.87\text{ B.M.}$)
Since Y gives 3 equivalents of $\text{AgCl}$ with $\text{AgNO}_3$, statement (d) is incorrect.
22. Hydrogen peroxide oxidises $[\text{Fe(CN)}_6]^{4-}$ to $[\text{Fe(CN)}_6]^{3-}$ in acidic medium but reduces $[\text{Fe(CN)}_6]^{3-}$ to $[\text{Fe(CN)}_6]^{4-}$ in alkaline medium. The other products formed are, respectively. [JEE Main 2018]
a. $\text{H}_2\text{O}$ and $(\text{H}_2\text{O} + \text{OH}^-)$
b. $(\text{H}_2\text{O} + \text{O}_2)$ and $\text{H}_2\text{O}$
c. $(\text{H}_2\text{O} + \text{O}_2)$ and $(\text{H}_2\text{O} + \text{OH}^-)$
d. $\text{H}_2\text{O}$ and $(\text{H}_2\text{O} + \text{O}_2)$
Answer: (d)
Solution:
In acidic medium: $2[\text{Fe(CN)}_6]^{4-} + \text{H}_2\text{O}_2 + 2\text{H}^+ \longrightarrow 2[\text{Fe(CN)}_6]^{3-} + 2\text{H}_2\text{O}$
In alkaline medium: $2[\text{Fe(CN)}_6]^{3-} + \text{H}_2\text{O}_2 + 2\text{OH}^- \longrightarrow 2[\text{Fe(CN)}_6]^{4-} + 2\text{H}_2\text{O} + \text{O}_2$
23. Diborane ($\text{B}_2\text{H}_6$) reacts independently with $\text{O}_2$ and $\text{H}_2\text{O}$ to produce, respectively [JEE Main 2019]
a. $\text{HBO}_2$ and $\text{H}_3\text{BO}_3$
b. $\text{H}_3\text{BO}_3$ and $\text{B}_2\text{O}_3$
c. $\text{B}_2\text{O}_3$ and $\text{H}_3\text{BO}_3$
d. $\text{B}_2\text{O}_3$ and $[\text{BH}_4]^-$
Answer: (c)
Solution:
$$\text{B}_2\text{H}_6 + 3\text{O}_2 \longrightarrow \text{B}_2\text{O}_3 + 3\text{H}_2\text{O}$$ $$\text{B}_2\text{H}_6 + 6\text{H}_2\text{O} \longrightarrow 2\text{H}_3\text{BO}_3 + 6\text{H}_2$$24. The correct order of the spin-only magnetic moment of metal ions in the following low spin complexes, $[\text{V(CN)}_6]^{4-}$, $[\text{Fe(CN)}_6]^{4-}$, $[\text{Ru}(\text{NH}_3)_6]^{3+}$, and $[\text{Cr}(\text{NH}_3)_6]^{2+}$, is : [JEE Main 2019]
a. $\text{V}^{2+} > \text{Cr}^{2+} > \text{Ru}^{3+} > \text{Fe}^{2+}$
b. $\text{V}^{2+} > \text{Ru}^{3+} > \text{Cr}^{2+} > \text{Fe}^{2+}$
c. $\text{Cr}^{2+} > \text{V}^{2+} > \text{Ru}^{3+} > \text{Fe}^{2+}$
d. $\text{Cr}^{2+} > \text{Ru}^{3+} > \text{Fe}^{2+} > \text{V}^{2+}$
Answer: (b)
Solution:
1. $[\text{V(CN)}_6]^{4-}: \text{V}^{2+} (3\text{d}^3) \implies t_{2g}^3 e_g^0 \implies n = 3$
2. $[\text{Ru}(\text{NH}_3)_6]^{3+}: \text{Ru}^{3+} (4\text{d}^5) \implies t_{2g}^5 e_g^0 \implies n = 1$
3. $[\text{Cr}(\text{NH}_3)_6]^{2+}: \text{Cr}^{2+} (3\text{d}^4) \implies t_{2g}^4 e_g^0 \implies n = 2$
4. $[\text{Fe(CN)}_6]^{4-}: \text{Fe}^{2+} (3\text{d}^6) \implies t_{2g}^6 e_g^0 \implies n = 0$
Order of spin-only magnetic moment: $\text{V}^{2+} (n=3) > \text{Cr}^{2+} (n=2) > \text{Ru}^{3+} (n=1) > \text{Fe}^{2+} (n=0)$
25. The ion that has $\text{sp}^3\text{d}^2$ hybridization for the central atom, is : [JEE Main 2019]
a. $[\text{ICl}_2]^-$
b. $[\text{IF}_6]^-$
c. $[\text{ICl}_4]^-$
d. $[\text{BrF}_2]^-$
Answer: (c)
Solution:
In $[\text{ICl}_4]^-$, central iodine atom has $7 + 4 + 1 = 12$ valence electrons, forming 4 bond pairs and 2 lone pairs.
$$\text{Steric Number} = 4 + 2 = 6 \implies \text{sp}^3\text{d}^2 \text{ hybridization (square planar)}.$$26. The calculated spin-only magnetic moments (BM) of the anionic and cationic species of $[\text{Fe}(\text{H}_2\text{O})_6]_2 [\text{Fe(CN)}_6]$, respectively, are : [JEE Main 2019]
a. 4.9 and 0
b. 2.84 and 5.92
c. 0 and 4.9
d. 0 and 5.92
Answer: (c)
Solution:
Anionic species: $[\text{Fe(CN)}_6]^{4-} \implies \text{Fe}^{2+} (3\text{d}^6)$, strong field ligand $\implies n = 0 \implies \mu = 0\text{ B.M.}$
Cationic species: $[\text{Fe}(\text{H}_2\text{O})_6]^{2+} \implies \text{Fe}^{2+} (3\text{d}^6)$, weak field ligand $\implies n = 4 \implies \mu = \sqrt{24} \approx 4.9\text{ B.M.}$
27. Among the following molecules / ions, $\text{C}_2^{2-}, \text{N}_2^{2-}, \text{O}_2^{2-}, \text{O}_2$ which one is diamagnetic and has the shortest bond length? [JEE Main 2019]
a. $\text{C}_2^{2-}$
b. $\text{N}_2^{2-}$
c. $\text{O}_2$
d. $\text{O}_2^{2-}$
Answer: (a)
Solution:
$\text{C}_2^{2-}$ has 14 electrons (isoelectronic with $\text{N}_2$).
Electronic configuration: $\sigma1\text{s}^2 \sigma^*1\text{s}^2 \sigma2\text{s}^2 \sigma^*2\text{s}^2 \pi2\text{p}_x^2 \pi2\text{p}_y^2 \sigma2\text{p}_z^2$
All electrons are paired $\implies$ Diamagnetic.
$\text{Bond Order} = \frac{10 - 4}{2} = 3$ (highest bond order $\implies$ shortest bond length).
28. The correct statement about $\text{ICl}_5$ and $\text{ICl}_4^-$ is [JEE Main 2019]
a. $\text{ICl}_5$ is trigonal bipyramidal and $\text{ICl}_4^-$ is tetrahedral.
b. $\text{ICl}_5$ is square pyramidal and $\text{ICl}_4^-$ is tetrahedral.
c. $\text{ICl}_5$ is square pyramidal and $\text{ICl}_4^-$ is square planar.
d. Both are isostructural.
Answer: (c)
Solution:
$\text{ICl}_5$: 5 bond pairs + 1 lone pair = Steric number 6 $\implies$ Square pyramidal.
$\text{ICl}_4^-$: 4 bond pairs + 2 lone pairs = Steric number 6 $\implies$ Square planar.
29. Two complexes $[\text{Cr}(\text{H}_2\text{O})_6]\text{Cl}_3$ (A) and $[\text{Cr}(\text{NH}_3)_6]\text{Cl}_3$ (B) are violet and yellow coloured, respectively. The incorrect statement regarding them is: [JEE Main 2019]
a. $\Delta_o$ value of (A) is less than that (B).
b. $\Delta_o$ value of (A) and (B) are calculated from the energies of violet and yellow light, respectively.
c. Both absorb energies corresponding to their complementary colors.
d. Both are paramagnetic with three unpaired electrons.
Answer: (b)
Solution:
The crystal field splitting energy ($\Delta_o$) corresponds to the light absorbed by the complex, which is of the complementary color to the observed/transmitted color (not calculated directly from the transmitted violet and yellow light colors).
30. The complex that has highest crystal field splitting energy ($\Delta_o$), is [JEE Main 2019]
a. $\text{K}_3[\text{Co(CN)}_6]$
b. $[\text{Co}(\text{NH}_3)_5(\text{H}_2\text{O})]\text{Cl}_3$
c. $\text{K}_2[\text{CoCl}_4]$
d. $[\text{Co}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2$
Answer: (a)
Solution:
According to spectrochemical series, $\text{CN}^-$ is the strongest field ligand among the given choices, producing the maximum crystal field splitting energy ($\Delta_o$).
Multiple Choice Questions
One and More than One Correct
31. If the bond length of CO bond in carbon monoxide is $1.128\text{ \AA}$, then what is the value of CO bond length in $\text{Fe(CO)}_5$? [JEE 2006]
a. $1.15\text{ \AA}$
b. $1.128\text{ \AA}$
c. $1.72\text{ \AA}$
d. $1.118\text{ \AA}$
Answer: (a)
Solution:
In metal carbonyls like $\text{Fe(CO)}_5$, back-bonding occurs where electron density is donated from the filled metal $d$-orbitals into the empty $\pi^*$ antibonding orbitals of $\text{CO}$. This synergic bonding weakens the $\text{C}\equiv\text{O}$ bond, increasing the $\text{C-O}$ bond length compared to free carbon monoxide ($1.128\text{ \AA}$). Therefore, the $\text{C-O}$ bond length becomes $1.15\text{ \AA}$.
32. Among the following metal carbonyls, the $\text{C-O}$ bond order is lowest in [JEE 2007 P-II]
a. $[\text{Mn(CO)}_6]^+$
b. $[\text{Fe(CO)}_5]$
c. $[\text{Cr(CO)}_6]$
d. $[\text{V(CO)}_6]^-$
Answer: (d)
Solution:
Greater the negative charge on the central metal ion, higher is the $d\pi-p\pi$ back-bonding capability into the $\pi^*$ orbital of $\text{CO}$. Higher back-donation weakens the $\text{C-O}$ bond and reduces its bond order. Since $[\text{V(CO)}_6]^-$ carries a net negative charge (metal in $-1$ oxidation state), it exhibits maximum back-bonding and thus the lowest $\text{C-O}$ bond order.
33. Both $\text{Ni(CO)}_4$ and $[\text{Ni(CN)}_4]^{2-}$ are diamagnetic. The hybridization of nickel in these complexes, respectively, are [JEE 2008 P-II]
a. $sp^3, sp^3$
b. $sp^3, dsp^2$
c. $dsp^2, sp^3$
d. $dsp^2, dsp^2$
Answer: (b)
Solution:
In $\text{Ni(CO)}_4$, $\text{Ni}$ is in $0$ oxidation state ($\text{3d}^8\text{4s}^2$). Strong ligand $\text{CO}$ pairs up electrons into $\text{3d}^{10}$, leading to $sp^3$ hybridization (tetrahedral).
In $[\text{Ni(CN)}_4]^{2-}$, $\text{Ni}^{2+}$ has $\text{3d}^8$ configuration. Strong field ligand $\text{CN}^-$ forces pairing within $3d$ orbitals, leaving one $3d$ orbital vacant for $dsp^2$ hybridization (square planar).
34. The spin only magnetic moment value (in Bohr magneton units) of $[\text{Cr(H}_2\text{O)}_6]^{2+}$ is [JEE 2009 P-II]
a. 0
b. 2.84
c. 4.90
d. 5.92
Answer: (c)
Solution:
$\text{Cr}^{2+}$ has an electronic configuration of $[\text{Ar}]3d^4$. Since $\text{H}_2\text{O}$ is a weak field ligand, no pairing occurs ($t_{2g}^3 e_g^1$), giving $n = 4$ unpaired electrons.
$$\mu = \sqrt{n(n+2)} = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\text{ B.M.}$$35. The correct structure of ethylenediaminetetraacetic acid (EDTA) is [JEE 2010 P-I]
a.
b.
c.
d.
Answer: (c)
Solution:
EDTA ($\text{ethylenediaminetetraacetic acid}$) contains two nitrogen atoms linked by an ethylene group, with each nitrogen attached to two acetate groups: $(\text{HOOC-CH}_2)_2\text{N-CH}_2\text{-CH}_2\text{-N}(\text{CH}_2\text{-COOH})_2$.
36. The ionization isomer of $[\text{Cr(H}_2\text{O)}_4\text{Cl(NO}_2)]\text{Cl}$ is [JEE 2010 P-I]
a. $[\text{Cr(H}_2\text{O)}_4(\text{O}_2\text{N})]\text{Cl}_2$
b. $[\text{Cr(H}_2\text{O)}_4\text{Cl}_2](\text{NO}_2)$
c. $[\text{Cr(H}_2\text{O)}_4\text{Cl(ONO)}]\text{Cl}$
d. $[\text{Cr(H}_2\text{O)}_4\text{Cl}_2(\text{NO}_2)]\cdot\text{H}_2\text{O}$
Answer: (b)
Solution:
Ionization isomers exchange ions inside and outside the coordination sphere. Interchanging $\text{Cl}^-$ outside with $\text{NO}_2^-$ inside yields $[\text{Cr(H}_2\text{O)}_4\text{Cl}_2](\text{NO}_2)$.
37. The complex showing a spin-only magnetic moment of 2.82 B.M. is [JEE 2010 P-II]
a. $\text{Ni(CO)}_4$
b. $[\text{NiCl}_4]^{2-}$
c. $\text{Ni(PPh}_3)_4$
d. $[\text{Ni(CN)}_4]^{2-}$
Answer: (b)
Solution:
A magnetic moment of $2.82\text{ B.M.}$ corresponds to $n = 2$ unpaired electrons ($\sqrt{2(2+2)} = \sqrt{8} \approx 2.83\text{ B.M.}$).
In $[\text{NiCl}_4]^{2-}$, $\text{Ni}^{2+}$ is $3d^8$. Weak ligand $\text{Cl}^-$ gives tetrahedral geometry with 2 unpaired electrons.
38. Geometrical shapes of the complexes formed by the reaction of $\text{Ni}^{2+}$ with $\text{Cl}^-$, $\text{CN}^-$ and $\text{H}_2\text{O}$ respectively, are [JEE 2011 P-I]
a. octahedral, tetrahedral and square planar
b. tetrahedral, square planar and octahedral
c. square planar, tetrahedral and octahedral
d. octahedral, square planar and octahedral
Answer: (b)
Solution:
1. Reaction with $\text{Cl}^-$ forms $[\text{NiCl}_4]^{2-}$ (Tetrahedral, $sp^3$).
2. Reaction with $\text{CN}^-$ forms $[\text{Ni(CN)}_4]^{2-}$ (Square planar, $dsp^2$).
3. Reaction with $\text{H}_2\text{O}$ forms $[\text{Ni(H}_2\text{O)}_6]^{2+}$ (Octahedral, $sp^3d^2$).
39. Among the following complexes (K–P):
$\text{K}_3[\text{Fe(CN)}_6]\text{ (K)}$, $[\text{Co(NH}_3)_6]\text{Cl}_3\text{ (L)}$, $\text{Na}_3[\text{Co(oxalate)}_3]\text{ (M)}$, $[\text{Ni(H}_2\text{O)}_6]\text{Cl}_2\text{ (N)}$, $\text{K}_2[\text{Pt(CN)}_4]\text{ (O)}$, $[\text{Zn(H}_2\text{O)}_6](\text{NO}_3)_2\text{ (P)}$
Following compounds are diamagnetic: [JEE 2011 P-II]
a. K, L, M, N
b. K, M, O, P
c. L, M, O, P
d. L, M, N, O
Answer: (c)
Solution:
$\text{L: } [\text{Co(NH}_3)_6]^{3+} \rightarrow \text{Co}^{3+} (3d^6)$, low spin, diamagnetic ($n=0$).
$\text{M: } [\text{Co(ox)}_3]^{3-} \rightarrow \text{Co}^{3+} (3d^6)$, low spin, diamagnetic ($n=0$).
$\text{O: } [\text{Pt(CN)}_4]^{2-} \rightarrow \text{Pt}^{2+} (5d^8)$, square planar, diamagnetic ($n=0$).
$\text{P: } [\text{Zn(H}_2\text{O)}_6]^{2+} \rightarrow \text{Zn}^{2+} (3d^{10})$, fully filled $d$-subshell, diamagnetic ($n=0$).
40. $\text{NiCl}_2\{\text{P}(\text{C}_2\text{H}_5)_2(\text{C}_6\text{H}_5)\}_2$ exhibits temperature dependent magnetic behavior (paramagnetic / diamagnetic). The coordination geometries of $\text{Ni}^{2+}$ in the paramagnetic and diamagnetic states are respectively: [JEE 2012 P-II]
a. tetrahedral and tetrahedral
b. square planar and square planar
c. tetrahedral and square planar
d. square planar and tetrahedral
Answer: (c)
Solution:
For $\text{Ni}^{2+}$ ($3d^8$): Paramagnetic state corresponds to tetrahedral geometry ($sp^3$, $n=2$ unpaired electrons), while diamagnetic state corresponds to square planar geometry ($dsp^2$, $n=0$ unpaired electrons).
41. Consider the following complex ions, P, Q and R:
$\text{P} = [\text{FeF}_6]^{3-}$, $\text{Q} = [\text{V(H}_2\text{O)}_6]^{2+}$, $\text{R} = [\text{Fe(H}_2\text{O)}_6]^{2+}$.
The correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.) is [JEE Adv. 2013 P-I]
a. R < Q < P
b. Q < R < P
c. R < P < Q
d. Q < P < R
Answer: (b)
Solution:
$\text{P: } [\text{FeF}_6]^{3-} \rightarrow \text{Fe}^{3+} (3d^5)$, weak field $\text{F}^- \rightarrow n = 5 \implies \mu = 5.92\text{ B.M.}$
$\text{Q: } [\text{V(H}_2\text{O)}_6]^{2+} \rightarrow \text{V}^{2+} (3d^3) \rightarrow n = 3 \implies \mu = 3.87\text{ B.M.}$
$\text{R: } [\text{Fe(H}_2\text{O)}_6]^{2+} \rightarrow \text{Fe}^{2+} (3d^6)$, weak field $\text{H}_2\text{O} \rightarrow n = 4 \implies \mu = 4.90\text{ B.M.}$
Order: $\text{Q } (3.87) < \text{R } (4.90) < \text{P } (5.92)$.
42. Which of the following complex species is not expected to exhibit optical isomerism? [JEE Main 2013]
a. $[\text{Co(en)}_3]^{3+}$
b. $[\text{Co(en)}_2\text{Cl}_2]^+$
c. $[\text{Co(NH}_3)_3\text{Cl}_3]$
d. $[\text{Co(en)(NH}_3)_2\text{Cl}_2]^+$
Answer: (c)
Solution:
$[\text{Co(NH}_3)_3\text{Cl}_3]$ is an $\text{MA}_3\text{B}_3$ type complex, existing as facial ($fac$) and meridional ($mer$) isomers. Both forms contain a plane of symmetry ($\sigma$) and are optically inactive (achiral).
43. The number of geometric isomers that can exist for square planar $[\text{Pt(Cl)(py)(NH}_3)(\text{NH}_2\text{OH})]^+ $ is (py = pyridine): [JEE Main 2015]
a. 2
b. 3
c. 4
d. 6
Answer: (b)
Solution:
A square planar complex of the type $[\text{Mabcd}]$ with four different monodentate ligands possesses 3 geometrical isomers (fixing one ligand and varying the ligand trans to it).
44. The pair having the same magnetic moment is: [At. No.: Cr = 24, Mn = 25, Fe = 26, Co = 27] [JEE Main 2016]
a. $[\text{Cr(H}_2\text{O)}_6]^{2+}$ and $[\text{Fe(H}_2\text{O)}_6]^{2+}$
b. $[\text{Mn(H}_2\text{O)}_6]^{2+}$ and $[\text{Cr(H}_2\text{O)}_6]^{2+}$
c. $[\text{CoCl}_4]^{2-}$ and $[\text{Fe(H}_2\text{O)}_6]^{2+}$
d. $[\text{Cr(H}_2\text{O)}_6]^{2+}$ and $[\text{CoCl}_4]^{2-}$
Answer: (a)
Solution:
In $[\text{Cr(H}_2\text{O)}_6]^{2+}$, $\text{Cr}^{2+}$ is $3d^4$ (weak ligand $\implies n = 4$ unpaired electrons).
In $[\text{Fe(H}_2\text{O)}_6]^{2+}$, $\text{Fe}^{2+}$ is $3d^6$ (weak ligand $\implies t_{2g}^4 e_g^2$, $n = 4$ unpaired electrons).
Both have $n = 4$ unpaired electrons, so they have the same spin-only magnetic moment ($\mu = 3.89\text{ B.M.}$ or $\sqrt{24} \approx 4.90\text{ B.M.}$).
45. Which one of the following complexes shows optical isomerism? [JEE Main 2016]
a. $cis-[\text{Co(en)}_2\text{Cl}_2]\text{Cl}$
b. $trans-[\text{Co(en)}_2\text{Cl}_2]\text{Cl}$
c. $[\text{Co(NH}_3)_4\text{Cl}_2]\text{Cl}$
d. $[\text{Co(NH}_3)_3\text{Cl}_3]$
Answer: (a)
Solution:
$cis-[\text{Co(en)}_2\text{Cl}_2]^+$ lacks any plane or center of symmetry, making it non-superimposable on its mirror image, hence optically active.
46. Among $\text{Ni(CO)}_4$, $[\text{NiCl}_4]^{2-}$, $[\text{Co(NH}_3)_4\text{Cl}_2]\text{Cl}$, $\text{Na}_3[\text{CoF}_6]$, $\text{Na}_2\text{O}_2$ and $\text{CsO}_2$, the total number of paramagnetic compounds is [JEE Adv. 2016 P-I]
a. 2
b. 3
c. 4
d. 5
Answer: (c)
Solution:
1. $\text{Ni(CO)}_4 \rightarrow \text{Ni}^0 (3d^{10}) \implies$ Diamagnetic.
2. $[\text{NiCl}_4]^{2-} \rightarrow \text{Ni}^{2+} (3d^8) \implies$ Paramagnetic ($2$ unpaired $e^-$).
3. $[\text{Co(NH}_3)_4\text{Cl}_2]^+ \rightarrow \text{Co}^{3+} (3d^6$, low spin) $\implies$ Diamagnetic.
4. $[\text{CoF}_6]^{3-} \rightarrow \text{Co}^{3+} (3d^6$, weak field $\text{F}^-) \implies$ Paramagnetic ($4$ unpaired $e^-$).
5. $\text{Na}_2\text{O}_2 \rightarrow \text{O}_2^{2-}$ (peroxide) $\implies$ Diamagnetic.
6. $\text{CsO}_2 \rightarrow \text{O}_2^-$ (superoxide, 1 unpaired electron) $\implies$ Paramagnetic.
Total paramagnetic compounds = 3 ($[\text{NiCl}_4]^{2-}$, $\text{Na}_3[\text{CoF}_6]$, $\text{CsO}_2$).
47. The geometries of the ammonia complexes of $\text{Ni}^{2+}$, $\text{Pt}^{2+}$ & $\text{Zn}^{2+}$, respectively, are [JEE Adv. 2016 P-II]
a. octahedral, square planar and tetrahedral
b. square planar, octahedral and tetrahedral
c. tetrahedral, square planar and octahedral
d. octahedral, tetrahedral and square planar
Answer: (a)
Solution:
1. $[\text{Ni(NH}_3)_6]^{2+} \rightarrow$ Octahedral ($sp^3d^2$).
2. $[\text{Pt(NH}_3)_4]^{2+} \rightarrow$ Square planar ($dsp^2$, $5d$ metal complexes are square planar).
3. $[\text{Zn(NH}_3)_4]^{2+} \rightarrow$ Tetrahedral ($sp^3$, $3d^{10}$ filled).
48. The compound(s) that exhibit(s) geometrical isomerism is(are) [JEE 2009 P-I]
a. $[\text{Pt(en)Cl}_2]$
b. $[\text{Pt(en)}_2]\text{Cl}_2$
c. $[\text{Pt(en)}_2\text{Cl}_2]\text{Cl}_2$
d. $[\text{Pt(NH}_3)_2\text{Cl}_2]$
Answer: (c, d)
Solution:
1. $[\text{Pt(en)}_2\text{Cl}_2]^{2+}$ is an octahedral complex of type $[\text{M(en)}_2\text{X}_2]$ which exhibits $cis$ and $trans$ geometrical isomerism.
2. $[\text{Pt(NH}_3)_2\text{Cl}_2]$ is a square planar complex of type $[\text{MA}_2\text{B}_2]$ which exhibits $cis$ and $trans$ geometrical isomerism.
49. The pair(s) of coordination complexes/ions exhibiting the same kind of isomerism is(are) [JEE Adv. 2013 P-I]
a. $[\text{Cr(NH}_3)_5\text{Cl}]\text{Cl}_2$ and $[\text{Cr(NH}_3)_4\text{Cl}_2]\text{Cl}$
b. $[\text{Co(NH}_3)_4\text{Cl}_2]^+$ and $[\text{Pt(NH}_3)_2(\text{H}_2\text{O})\text{Cl}]^+$
c. $[\text{CoBr}_2\text{Cl}_2]^{2-}$ and $[\text{PtBr}_2\text{Cl}_2]^{2-}$
d. $[\text{Pt(NH}_3)_3(\text{NO}_2)]\text{Cl}$ and $[\text{Pt(NH}_3)_3\text{Cl}]\text{Br}$
Answer: (b, d)
Solution:
Option b: $[\text{Co(NH}_3)_4\text{Cl}_2]^+$ (octahedral) and $[\text{Pt(NH}_3)_2(\text{H}_2\text{O})\text{Cl}]^+$ (square planar) both exhibit geometrical isomerism.
Option d: $[\text{Pt(NH}_3)_3(\text{NO}_2)]\text{Cl}$ and $[\text{Pt(NH}_3)_3\text{Cl}]\text{Br}$ both exhibit ionization isomerism.
50. On treatment of 100 mL of 0.1 M solution of $\text{CoCl}_3\cdot 6\text{H}_2\text{O}$ with excess $\text{AgNO}_3$, $1.2 \times 10^{22}$ ions are precipitated. The complex is [JEE Main 2017]
a. $[\text{Co(H}_2\text{O)}_5\text{Cl}]\text{Cl}_2\cdot\text{H}_2\text{O}$
b. $[\text{Co(H}_2\text{O)}_4\text{Cl}_2]\text{Cl}\cdot 2\text{H}_2\text{O}$
c. $[\text{Co(H}_2\text{O)}_3\text{Cl}_3]\cdot 3\text{H}_2\text{O}$
d. $[\text{Co(H}_2\text{O)}_6]\text{Cl}_3$
Answer: (a)
Solution:
$$\text{Moles of complex} = \frac{100}{1000} \times 0.1 = 0.01\text{ mol}$$ $$\text{Moles of precipitated }\text{Cl}^- = \frac{1.2 \times 10^{22}}{6.022 \times 10^{23}} \approx 0.02\text{ mol}$$ $$\text{Number of ionizable }\text{Cl}^- \text{ ions per complex molecule} = \frac{0.02}{0.01} = 2$$Therefore, 2 $\text{Cl}^-$ ions lie outside the coordination sphere: $[\text{Co(H}_2\text{O)}_5\text{Cl}]\text{Cl}_2\cdot\text{H}_2\text{O}$.
51. Addition of excess aqueous ammonia to a pink coloured aqueous solution of $\text{MCl}_2\cdot 6\text{H}_2\text{O}$ (X) and $\text{NH}_4\text{Cl}$ gives an octahedral complex Y in the presence of air. In aqueous solution, complex Y behaves as 1 : 3 electrolyte. The reaction of X with excess HCl at room temperature results in the formation of a blue coloured complex Z. The calculated spin only magnetic moment of X and Z is 3.87 B.M., whereas it is zero for complex Y. Among the following options, which statement(s) is(are) correct? [JEE Adv. 2017 P-I]
a. The hybridization of the central metal ion in Y is $d^2sp^3$
b. When X and Z are in equilibrium at 0°C, the colour of the solution is pink
c. Z is a tetrahedral complex
d. Addition of silver nitrate to Y gives only two equivalents of silver chloride
Answer: (a, b, c)
Solution:
$\text{X} = [\text{Co(H}_2\text{O)}_6]\text{Cl}_2$ (pink, $\text{Co}^{2+} 3d^7$, $\mu = 3.87\text{ B.M.}$)
$\text{Y} = [\text{Co(NH}_3)_6]\text{Cl}_3$ (1:3 electrolyte, $\text{Co}^{3+} 3d^6$ low spin, diamagnetic $\mu=0$, $d^2sp^3$ hybridization). $\text{AgNO}_3$ precipitates 3 equivalents of $\text{AgCl}$.
$\text{Z} = [\text{CoCl}_4]^{2-}$ (blue tetrahedral complex, $\mu = 3.87\text{ B.M.}$)
At low temperature (0°C), the equilibrium $[\text{Co(H}_2\text{O)}_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CoCl}_4]^{2-} + 6\text{H}_2\text{O}$ shifts to the left, favoring the pink $[\text{Co(H}_2\text{O)}_6]^{2+}$.
52. Hydrogen peroxide oxidises $[\text{Fe(CN)}_6]^{4-}$ to $[\text{Fe(CN)}_6]^{3-}$ in acidic medium but reduces $[\text{Fe(CN)}_6]^{3-}$ to $[\text{Fe(CN)}_6]^{4-}$ in alkaline medium. The other products formed are, respectively: [JEE Main 2018]
a. $\text{H}_2\text{O}$ and $(\text{H}_2\text{O} + \text{OH}^-)$
b. $(\text{H}_2\text{O} + \text{O}_2)$ and $\text{H}_2\text{O}$
c. $(\text{H}_2\text{O} + \text{O}_2)$ and $(\text{H}_2\text{O} + \text{OH}^-)$
d. $\text{H}_2\text{O}$ and $(\text{H}_2\text{O} + \text{O}_2)$
Answer: (d)
Solution:
In acidic medium (oxidizing action of $\text{H}_2\text{O}_2$):
$$2[\text{Fe(CN)}_6]^{4-} + \text{H}_2\text{O}_2 + 2\text{H}^+ \longrightarrow 2[\text{Fe(CN)}_6]^{3-} + 2\text{H}_2\text{O}$$In alkaline medium (reducing action of $\text{H}_2\text{O}_2$):
$$2[\text{Fe(CN)}_6]^{3-} + \text{H}_2\text{O}_2 + 2\text{OH}^- \longrightarrow 2[\text{Fe(CN)}_6]^{4-} + 2\text{H}_2\text{O} + \text{O}_2$$53. Diborane ($\text{B}_2\text{H}_6$) reacts independently with $\text{O}_2$ and $\text{H}_2\text{O}$ to produce, respectively [JEE Main 2019]
a. $\text{HBO}_2$ and $\text{H}_3\text{BO}_3$
b. $\text{H}_3\text{BO}_3$ and $\text{B}_2\text{O}_3$
c. $\text{B}_2\text{O}_3$ and $\text{H}_3\text{BO}_3$
d. $\text{B}_2\text{O}_3$ and $[\text{BH}_4]^-$
Answer: (c)
Solution:
$$\text{B}_2\text{H}_6 + 3\text{O}_2 \longrightarrow \text{B}_2\text{O}_3 + 3\text{H}_2\text{O}$$ $$\text{B}_2\text{H}_6 + 6\text{H}_2\text{O} \longrightarrow 2\text{H}_3\text{BO}_3 + 6\text{H}_2$$54. The correct order of the spin-only magnetic moment of metal ions in the following low spin complexes, $[\text{V(CN)}_6]^{4-}$, $[\text{Fe(CN)}_6]^{4-}$, $[\text{Ru(NH}_3)_6]^{3+}$, and $[\text{Cr(NH}_3)_6]^{2+}$, is : [JEE Main 2019]
a. $\text{V}^{2+} > \text{Cr}^{2+} > \text{Ru}^{3+} > \text{Fe}^{2+}$
b. $\text{V}^{2+} > \text{Ru}^{3+} > \text{Cr}^{2+} > \text{Fe}^{2+}$
c. $\text{Cr}^{2+} > \text{V}^{2+} > \text{Ru}^{3+} > \text{Fe}^{2+}$
d. $\text{Cr}^{2+} > \text{Ru}^{3+} > \text{Fe}^{2+} > \text{V}^{2+}$
Answer: (b)
Solution:
All are low spin complexes:
1. $[\text{V(CN)}_6]^{4-} \rightarrow \text{V}^{2+} (t_{2g}^3 e_g^0) \implies 3\text{ unpaired } e^-$.
2. $[\text{Ru(NH}_3)_6]^{3+} \rightarrow \text{Ru}^{3+} (4d^5, t_{2g}^5 e_g^0) \implies 1\text{ unpaired } e^-$.
3. $[\text{Cr(NH}_3)_6]^{2+} \rightarrow \text{Cr}^{2+} (3d^4, t_{2g}^4 e_g^0) \implies 2\text{ unpaired } e^-$.
4. $[\text{Fe(CN)}_6]^{4-} \rightarrow \text{Fe}^{2+} (3d^6, t_{2g}^6 e_g^0) \implies 0\text{ unpaired } e^-$.
Order of magnetic moment: $\text{V}^{2+} (3) > \text{Cr}^{2+} (2) > \text{Ru}^{3+} (1) > \text{Fe}^{2+} (0)$.
55. The ion that has $sp^3d^2$ hybridization for the central atom, is : [JEE Main 2019]
a. $[\text{ICl}_2]^-$
b. $[\text{IF}_6]^-$
c. $[\text{ICl}_4]^-$
d. $[\text{BrF}_2]^-$
Answer: (c)
Solution:
In $[\text{ICl}_4]^-$, central iodine has 7 valence electrons $+ 1$ negative charge $= 8$ electrons.
Forming $4$ $\sigma$-bonds with $\text{Cl}$ leaves $2$ lone pairs.
Steric Number $= 4\text{ bonds} + 2\text{ lone pairs} = 6 \implies sp^3d^2$ hybridization (square planar shape).
56. The calculated spin-only magnetic moments (BM) of the anionic and cationic species of $[\text{Fe(H}_2\text{O)}_6]_2[\text{Fe(CN)}_6]$, respectively, are : [JEE Main 2019]
a. 4.9 and 0
b. 2.84 and 5.92
c. 0 and 4.9
d. 0 and 5.92
Answer: (c)
Solution:
The compound dissociates into cationic species $[\text{Fe(H}_2\text{O)}_6]^{2+}$ and anionic species $[\text{Fe(CN)}_6]^{4-}$.
Anionic species $[\text{Fe(CN)}_6]^{4-}: \text{Fe}^{2+} (3d^6)$ in strong field ligand $\implies t_{2g}^6 e_g^0 \implies n = 0 \implies \mu = 0\text{ B.M.}$
Cationic species $[\text{Fe(H}_2\text{O)}_6]^{2+}: \text{Fe}^{2+} (3d^6)$ in weak field ligand $\implies t_{2g}^4 e_g^2 \implies n = 4 \implies \mu = \sqrt{24} \approx 4.9\text{ B.M.}$
57. Among the following molecules / ions, $\text{C}_2^{2-}$, $\text{N}_2^{2-}$, $\text{O}_2^{2-}$, $\text{O}_2$, which one is diamagnetic and has the shortest bond length? [JEE Main 2019]
a. $\text{C}_2^{2-}$
b. $\text{N}_2^{2-}$
c. $\text{O}_2$
d. $\text{O}_2^{2-}$
Answer: (a)
Solution:
$\text{C}_2^{2-}$ has $14$ electrons (isoelectronic with $\text{N}_2$).
Molecular orbital configuration: $\sigma1s^2 \sigma^*1s^2 \sigma2s^2 \sigma^*2s^2 \pi2p_x^2 = \pi2p_y^2 \sigma2p_z^2$.
All electrons are paired $\implies$ diamagnetic.
Bond order $= 3$ (highest bond order among given species $\implies$ shortest bond length).
58. The correct statement about $\text{ICl}_5$ and $\text{ICl}_4^-$ is [JEE Main 2019]
a. $\text{ICl}_5$ is trigonal bipyramidal and $\text{ICl}_4^-$ is tetrahedral.
b. $\text{ICl}_5$ is square pyramidal and $\text{ICl}_4^-$ is tetrahedral.
c. $\text{ICl}_5$ is square pyramidal and $\text{ICl}_4^-$ is square planar.
d. Both are isostructural.
Answer: (c)
Solution:
1. $\text{ICl}_5$: 5 bond pairs $+ 1$ lone pair $\implies$ Steric Number $= 6 \implies$ Square pyramidal geometry.
2. $\text{ICl}_4^-$: 4 bond pairs $+ 2$ lone pairs $\implies$ Steric Number $= 6 \implies$ Square planar geometry.
59. Two complexes $[\text{Cr(H}_2\text{O)}_6]\text{Cl}_3$ (A) and $[\text{Cr(NH}_3)_6]\text{Cl}_3$ (B) are violet and yellow coloured, respectively. The incorrect statement regarding them is: [JEE Main 2019]
a. $\Delta_0$ value of (A) is less than that of (B).
b. $\Delta_0$ value of (A) and (B) are calculated from the energies of violet and yellow light, respectively.
c. Both absorb energies corresponding to their complementary colors.
d. Both are paramagnetic with three unpaired electrons.
Answer: (b)
Solution:
The observed color of a complex is complementary to the wavelength of light absorbed. Therefore, crystal field splitting energy ($\Delta_0$) is calculated from the absorbed light frequency/wavelength, not directly from the wavelength of the transmitted (observed) light color.
60. The complex that has highest crystal field splitting energy ($\Delta_0$), is [JEE Main 2019]
a. $\text{K}_3[\text{Co(CN)}_6]$
b. $[\text{Co(NH}_3)_5(\text{H}_2\text{O})]\text{Cl}_3$
c. $\text{K}_2[\text{CoCl}_4]$
d. $[\text{Co(NH}_3)_5\text{Cl}]\text{Cl}_2$
Answer: (a)
Solution:
According to the spectrochemical series, $\text{CN}^-$ is a stronger field ligand than $\text{NH}_3$, $\text{H}_2\text{O}$, and $\text{Cl}^-$. A stronger ligand causes greater crystal field splitting ($\Delta_0$). Hence, $[\text{Co(CN)}_6]^{3-}$ has the highest $\Delta_0$.
61. The metal that forms nitride by reacting directly with $\text{N}_2$ of air, is [JEE Main 2019]
a. K
b. Cs
c. Li
d. Rb
Answer: (c)
Solution:
Due to its high lattice energy resulting from its small ionic size, lithium ($\text{Li}$) is the only alkali metal that reacts directly with atmospheric $\text{N}_2$ to form lithium nitride ($\text{Li}_3\text{N}$):
$$6\text{Li} + \text{N}_2 \longrightarrow 2\text{Li}_3\text{N}$$
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