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Coordination Compounds Set-8 MCQ Online Practice Test | Class 12 | Quiz Tweek

Coordination Compounds Set-8

Practice Important MCQs for Coordination Compounds — Chemistry, Class 12.

Coordination compounds are central to both CBSE and competitive exams because they link chemical bonding with real, observable properties such as colour, magnetism, stereochemistry (cis/trans and optical isomerism), and reactivity (kinetic stability and ligand effects). Mastering this chapter helps you solve problems quickly using oxidation-state counting, crystal-field theory (high-spin vs low-spin), and formation/ substitution equilibrium—core skills repeatedly tested across boards and entrance exams.

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Q1. Calculate the spin-only magnetic moment (in Bohr magneton) for the complex $[Fe(CN)_6]^{3-}$ , assuming CN $^-$ is a strong-field ligand and the complex is low-spin.

(A)

$2.83\ \mathrm{BM}$

(B)

$1.73\ \mathrm{BM}$

(C)

$5.92\ \mathrm{BM}$

(D)

$0.00\ \mathrm{BM}

Q2. How many stereoisomers (including enantiomers) are possible for the octahedral complex $[Co(en)_2Cl_2]^+$ (en = ethylenediamine)?

(A)

Two

(B)

Four

(C)

Three

(D)

Five

Q3. For the Ni(II) complexes $[Ni(CN)_4]^{2-}$ and $[NiCl_4]^{2-}$ , which statement correctly describes their magnetic behaviour and number of unpaired electrons?

(A)

$[Ni(CN)_4]^{2-}$ is diamagnetic (0 unpaired electrons), while $[NiCl_4]^{2-}$ is paramagnetic with two unpaired electrons (μ ≈ $2.83\ \mathrm{BM}$ ).

(B)

$[Ni(CN)_4]^{2-}$ is paramagnetic with two unpaired electrons, while $[NiCl_4]^{2-}$ is diamagnetic.

(C)

Both complexes are diamagnetic.

(D)

Both complexes are paramagnetic with one unpaired electron.

Q4. (A) Square-planar $d^8$ complexes are usually diamagnetic. (R) In a square-planar crystal field the $d_{x^2-y^2}$ orbital lies at much higher energy than the other $d$ orbitals, so it remains unoccupied and the remaining electrons pair up.

(A)

Both A and R are false.

(B)

A is true but R is false.

(C)

A is false but R is true.

(D)

Both A and R are true and R correctly explains A.

Q5. When aqueous Co(II) as $[Co(H_2O)_6]^{2+}$ (pink) is converted to $[CoCl_4]^{2-}$ (blue) on addition of excess Cl $^-$ (concentrated HCl), what is the expected change in magnetic moment and why?

(A)

It increases because tetrahedral $[CoCl_4]^{2-}$ is assumed to be high-spin with four unpaired electrons ( $\mu \approx 4.90\ \mathrm{BM}$ ).

(B)

It remains approximately unchanged because both octahedral high-spin and tetrahedral high-spin Co(II) ( $d^7$ ) species have three unpaired electrons (μ ≈ $3.87\ \mathrm{BM}$ ).

(C)

It decreases because chloride is assumed to cause low-spin pairing in tetrahedral $d^7$ (giving 1 unpaired electron, $\mu \approx 1.73\ \mathrm{BM}$ ).

(D)

It decreases because $d^7$ tetrahedral is assumed to have two unpaired electrons ( $\mu \approx 2.83\ \mathrm{BM}$ ).

Q6. Determine the oxidation state of cobalt in the coordination compound $[Co(NH_3)_4Cl_2]Br$ .

(A)

$+2$

(B)

$+1$

(C)

$+3$

(D)

$+4$

Q7. For the complexes $[Fe(H_2O)_6]^{2+}$ and $[Fe(CN)_6]^{4-}$ (both iron in same oxidation state), predict the number of unpaired electrons in each and calculate the spin-only magnetic moment of $[Fe(H_2O)_6]^{2+}$ .

(A)

$[Fe(H_2O)_6]^{2+}$ : 4 unpaired, $[Fe(CN)_6]^{4-}$ : 0 unpaired; $\mu_{\text{spin-only}}=\sqrt{4(4+2)}\approx4.90\ \mu_B$

(B)

$[Fe(H_2O)_6]^{2+}$ : 2 unpaired, $[Fe(CN)_6]^{4-}$ : 2 unpaired; $\mu_{\text{spin-only}}\approx2.83\ \mu_B$

(C)

$[Fe(H_2O)_6]^{2+}$ : 0 unpaired, $[Fe(CN)_6]^{4-}$ : 4 unpaired; $\mu_{\text{spin-only}}=0\ \mu_B$

(D)

$[Fe(H_2O)_6]^{2+}$ : 6 unpaired, $[Fe(CN)_6]^{4-}$ : 0 unpaired; $\mu_{\text{spin-only}}\approx\sqrt{6(6+2)}\ \mu_B$

Q8. Assertion: The complex $[Cr(H_2O)_6]^{3+}$ is kinetically inert while $[Ni(H_2O)_6]^{2+}$ is labile.
Reason: $Cr^{3+}$ has higher charge density and larger ligand–metal bond strength (and greater ligand-field stabilization) than $Ni^{2+}$ , which raises the activation energy for ligand substitution.

(A)

Both assertion and reason are true, but the reason is not the correct explanation of the assertion.

(B)

Both assertion and reason are false.

(C)

Assertion is true but reason is false.

(D)

Both assertion and reason are true and the reason correctly explains the assertion.

Q9. A solution initially contains $0.010\ \mathrm{M}$ $Cu^{2+}$ and $0.100\ \mathrm{M}$ $NH_3$ . The overall formation constant for $[Cu(NH_3)_4]^{2+}$ is $\beta_4=1\times10^{13}$ . Assuming only the tetraammine complex forms, estimate the equilibrium concentration of free $Cu^{2+}$ (take $[NH_3]_{\text{eq}}\approx0.06\ \mathrm{M}$ after complexation).

(A)

$\approx 1\times10^{-8}\ \mathrm{M}$

(B)

$\approx 7.7\times10^{-11}\ \mathrm{M}$

(C)

$\approx 1\times10^{-3}\ \mathrm{M}$

(D)

$\approx 1\times10^{-13}\ \mathrm{M}$

Q10. A cobalt–ammonia compound has formula $CoCl_3\cdot xNH_3$ . On treatment with excess $AgNO_3$ it precipitates exactly 2 moles of $AgCl$ per mole of the compound. Determine $x$ and the structural formula of the complex.

(A)

$x=5;\quad [Co(NH_3)_5Cl]Cl_2$

(B)

$x=6;\quad [Co(NH_3)_6]Cl_3$

(C)

$x=4;\quad [Co(NH_3)_4Cl_2]Cl$

(D)

$x=3;\quad [Co(NH_3)_3Cl_3]$

Q11. For the coordination compound $[\text{Cr}(\text{NH}_3)_4\text{Cl}_2]\text{Cl}$ , what are the oxidation state of Cr and the coordination number of the complex ion $[\text{Cr}(\text{NH}_3)_4\text{Cl}_2]^+$ ?

(A)

Oxidation state $+2$ , coordination number $6$

(B)

Oxidation state $+3$ , coordination number $6$

(C)

Oxidation state $+3$ , coordination number $4$

(D)

Oxidation state $+4$ , coordination number $6$

Q12. The complexes $[ \text{Fe(CN)}_6 ]^{4-}$ and $[ \text{Fe(H}_2\text{O)}_6 ]^{2+}$ contain Fe in the same oxidation state. How many unpaired electrons are present in $[ \text{Fe(CN)}_6 ]^{4-}$ and $[ \text{Fe(H}_2\text{O)}_6 ]^{2+}$ respectively?

(A)

2 and 4

(B)

4 and 0

(C)

0 and 2

(D)

0 and 4

Q13. A metal ion with electronic configuration $d^6$ forms octahedral complexes $[M(L_1)_6]$ and $[M(L_2)_6]$ . For $L_1$ , $\Delta_o=140\ \text{kJ mol}^{-1}$ ; for $L_2$ , $\Delta_o=80\ \text{kJ mol}^{-1}$ . The pairing energy $P$ for the metal is $100\ \text{kJ mol}^{-1}$ . Predict the spin states of $[M(L_1)_6]$ and $[M(L_2)_6]$ .

(A)

$[M(L_1)_6]$ is low-spin; $[M(L_2)_6]$ is high-spin

(B)

Both $[M(L_1)_6]$ and $[M(L_2)_6]$ are low-spin

(C)

Both $[M(L_1)_6]$ and $[M(L_2)_6]$ are high-spin

(D)

$[M(L_1)_6]$ is high-spin; $[M(L_2)_6]$ is low-spin

Q14. Assertion (A): Most stable complexes of $d^8$ ions (for example Ni(II), Pd(II), Pt(II)) adopt square-planar geometry and are diamagnetic.
Reason (R): Square-planar geometry occurs because the ligand field splitting in square-planar complexes is smaller than in octahedral complexes, which prevents pairing of electrons.

(A)

Both A and R are true, and R explains A

(B)

Both A and R are true, but R does not explain A

(C)

A is true but R is false

(D)

A is false but R is true

Q15. Which of the following Ni(II) complexes are expected to be paramagnetic (show unpaired electrons)?
(1) $[\text{NiCl}_4]^{2-}$ , (2) $[\text{Ni(CN)}_4]^{2-}$ , (3) $[\text{Ni(H}_2\text{O})_6]^{2+}$

(A)

Only (1)

(B)

(1) and (3) only

(C)

(2) and (3) only

(D)

All three (1), (2) and (3)

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