Coordination CompoundsNEET Chemistry · Class 12 · NCERT Chapter 5

Solution

The complex ${\text{[Fe(H}}_2{\text{O)}}_6{\text{]}}^{3+}$ has a ${\text{d}}^5$ configuration, which results in zero crystal field stabilization energy (CFSE) in an octahedral field due to the high-spin arrangement. NCERT XII chapter Coordination Compounds explains that in a high-spin ${\text{d}}^5$ complex, all five electrons occupy the five ${\text{t}}_{2g}$ and ${\text{e}}_g$ orbitals singly, leading to zero CFSE, so option (b) is correct.

Solution

The magnetic moment $\mu =\sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. For ${\text{Ni}}^{2+}$, the electronic configuration is $3d^8$, giving 2 unpaired electrons. Substituting, $\mu =\sqrt{2(2+2)}=\sqrt{8}=2.83\text{BM}$, so option (b) is correct.

Solution

Cisplatin, $\text{cis}-[{\text{PtCl}}_2({\text{NH}}_3{)}_2]$, is a well-known anticancer agent used in chemotherapy. NCERT XII chapter Coordination Compounds discusses its structure and biological activity, so option (b) is correct.

Solution

[Fe(CN)6]3− Hexacyanidoferrate (III) ion

Solution

[Ni(CN)4]2− Oxidation state of Ni is +2 x−4=2 x=+2

Solution

[M(en)2(C2O4)]Cl Oxidation state of M = +3 Coordination number of M = 6 Sum of oxidation state + Coordination number = 3 + 6 = 9

Solution

[Co(en)2Cl2]Cl Possible isomers (i) Geometrical isomers (ii) In trans form plane of symmetry present, so trans form is optically inactive but cis is optically active. Total number of stereoisomer = 2 + 1 = 3 www.vedantu.com 51

Solution

Metal carbon bond in metal carbonyls possess both σ and π character. M – C π bond is formed by donation of a pair of electrons from filled orbital of metal into vacant antibonding π orbital of CO. CO bond length increases if M has more tendency to donate lone pair by metal more CO bond length.

Solution

Complexes are respectively [Co(NH 3)6]Cl 3, [Co(NH3)5Cl]Cl2 and [Co(NH3)4Cl2]Cl

Solution

The order of the ligand in the spectrochemical series H2O < NH3 < en Hence, the wavelength of the light observed will be in the order [Co(H2O)6]3+ < [Co(NH3)6]3+ < [Co(en)3]3+ Thus, wavelength absorbed will be in the opposite order i.e., [Co(en)3]3+, [Co(NH3)6]3+, [Co(H2O)6]3+ www.vedantu.com 17

Solution

[Mn(CN)6]3– Mn(III) = [Ar]3d4 CN– being strong field ligand forces pairing of electrons This gives 40 2ggte ∴ Mn(III) = [Ar] 3d 4s 4p ds p 23 ∵Coordination number of Mn = 6 ∴ Structure = octahedral [Mn(CN)6]3– = [Ar] ds p 23 ×× ×× ×× ×× ×× ××

Solution

The complex $[{\text{CoCl}}_2(\text{en}{)}_2]$ can exhibit geometrical isomerism due to the arrangement of the bidentate ethylenediamine (en) ligands and the chloride ligands around the central cobalt ion. NCERT XII chapter Coordination Compounds explains that this type of complex can form cis and trans isomers, so option (c) is correct.

Solution

The complex ${\text{[Ni(CO)}}_4\text{]}$ has a tetrahedral geometry and is diamagnetic. Nickel in this complex is in the +2 oxidation state with a ${\text{d}}^8$ electron configuration, and the strong field ligands (CO) cause all electrons to pair up, resulting in a diamagnetic species, so option (b) is correct.

Solution

AgNO3+KI(EXCESS) → [𝐴𝑔𝐼]𝐼−

Solution

Urea reacts with water to form ammonia (A), which decomposes to form more ammonia (B). Ammonia forms a deep blue solution with ${\text{Cu}}^{2+}$ ions, resulting in the complex ion ${\text{[Cu(NH}}_3{\text{)}}_4{\text{]}}^{2+}$ (C). Therefore, option (b) is correct.

Solution

The correct order of increasing field strength of ligands is ${\text{SCN}}^{-}<{\text{F}}^{-}<\text{CO}<{\text{CN}}^{-}$. This follows the spectrochemical series, which ranks ligands based on their ability to split the d-orbitals of the central metal ion, as described in NCERT XII chapter Coordination Compounds. Thus, option (a) is correct.

Solution

EDTA is a hexadentate ligand with four oxygen and two nitrogen donor atoms. This is consistent with the structure of EDTA as described in NCERT XII chapter Coordination Compounds, making option (a) correct.

Solution

- [Fe(CN)${}_6$]${}^{3-}$: Low spin, ${\text{Fe}}^{3+}$, 1 unpaired electron, 1.73 BM.
- [Fe(H${}_2$O)${}_6$]${}^{3+}$: High spin, ${\text{Fe}}^{3+}$, 5 unpaired electrons, 0 BM.
- [Fe(CN)${}_6$]${}^{4-}$: Low spin, ${\text{Fe}}^{2+}$, 0 unpaired electrons, 4.90 BM.
- [Fe(H${}_2$O)${}_6$]${}^{2+}$: High spin, ${\text{Fe}}^{2+}$, 4 unpaired electrons, 5.92 BM.

Thus, the correct matches are (a)-(iv), (b)-(ii), (c)-(iii), (d)-(i), so option (b) is correct.

Solution

[Ag(H2O)2][Ag(CN)2] IUPAC name : diaquasilver(I)dicyanidoargentate(I)

Solution

Stronger the field strength of ligand, higher will be the energy absorbed by the complex. ⇒ ‘en’ has a stronger field strength than ‘H2O’ according to spectrochemical series ∴ Correct order of energy absorbed will be: [Ni(en)3]2+ > [Ni(H2O)2(en)2]2+ > [Ni(H2O)4(en)]2+ i.e. (C) > (A) > (B)

Solution

A homoleptic complex contains only one type of ligand. Potassium trioxalatoaluminate (III) has only oxalate ligands, making it homoleptic. NCERT XII chapter Coordination Compounds defines homoleptic complexes, so option (b) is correct.

Solution

In [Co(NH3)6]3+, Co3+ ion is having 3d6 configuration. Electronic configuration of Co3+ : In presence of NH3 ligand, pairing of electrons takes place and it becomes diamagnetic complex ion. In presence of NH3 ligand : ∴ [Co(NH3)6]3+ is octahedral with d2sp3 hybridisation and it is diamagnetic in nature. In case of [CoF6]3–, Co is in +3 oxidation state and it is having 3d6 configuration. In presence of weak field F– ligand, pairing does not take place. In presence of F– ligands : ∴ In [CoF6]3–, Co3+ is sp3d2 hybridised with four unpaired electrons, so it is paramagnetic in nature.

Solution

A. [Co(NH3)5(NO2)]Cl2 II. Linkage isomerism due to ‘N’ and ‘O’ linkage by NO2 B. [Co(NH3)5(SO4)]Br III. Ionization isomerism C. [Co(NH3)6][Cr(CN)6] IV. Coordination isomerism D. [Co(H2O)6]Cl3 I. Solvate isomerism

Solution

[Co(NH3)6]3+ is a homoleptic complex as only one type of ligands (NH 3) is coordinated with Co 3+ ion. While [Co(NH3)4Cl2]+ is a heteroleptic complex in which Co 3+ ion is ligated with more than one type of ligands, i.e., NH3 and Cl–. - 44 - NEET (UG)-2024 (Code-Q1)

Solution

A. $[NiC{l}_4{]}^{2-}$; $N{i}^{+2}$; $3d^8$; $s{p}^3$ hybridisation; 2 unpaired electrons; paramagnetic
B. $Ni(CO{)}_4$; Ni; $3d^{8}4s^2$; $s{p}^3$ hybridisation; Zero unpaired electron; diamagnetic
C. $[Ni(CN{)}_4{]}^{2-}$; $N{i}^{+2}$; $3d^8$; $ds{p}^2$ hybridisation; Zero unpaired electron; diamagnetic
D. $[Ni(H_{2}O{)}_6{]}^{2+}$; $N{i}^{2+}$, $3d^8$; $s{p}^3{d}^2$ hybridisation; Two unpaired electron; paramagnetic
E. $Ni(PP{h}_3{)}_4$; Ni; $3d^{8}4s^2$, $s{p}^3$ hybridisation; zero unpaired electron; Diamagnetic

Solution

Sol. $\lambda \propto \frac{1}{\text{strength of ligand}}$

$\lambda \propto \frac{1}{\text{splitting}}$

A. $[Co(N{H}_3{)}_6{]}^{3+}$ ; 475 nm
B. $[Co(CN{)}_6{]}^{3-}$ ; 310 nm
C. $[Cu(H_{2}O{)}_4{]}^{2+}$ ; 600 nm
D. $[Ti(H_{2}O{)}_6{]}^{3+}$ ; 498 nm

Order of $\lambda =C>D>A>B$

Solution

Conductance of any complex depends on the following factor.
(1) Number of ions produced by complex.
(2) If number of ions are same then we will check charge on complex unit.

(1) $\left[{\text{Co}}^{+3}({\text{NH}}_3{)}_3{\text{Cl}}_3\right]$
(2) $\left[{\text{Co}}^{+2}({\text{NH}}_3{)}_4{\text{Cl}}_2\right]$
Both complex units have no charge. Therefore both complex units have same conductance.

(3) [Co(NH${}_3$)${}_6$]Cl${}_3$ → [Co(NH${}_3$)${}_6$]${}^{+3}$ + 3Cl${}^{-}$

(4) [Co(NH${}_3$)${}_5$Cl]Cl → [Co(NH${}_3$)${}_5$Cl]${}^{+}$ + Cl${}^{-}$

Solution

- A (Coordination number 6) - I (Octahedral geometry)
- B (Coordination number 4) - II (Square planar geometry)
- C (Coordination number 4) - III (Tetrahedral geometry)
- D (Coordination number 2) - IV (Linear geometry)

Solution

A. Haber process — Fe catalyst (synthesis of NH₃ from N₂ + H₂). → I.
B. Wacker oxidation — PdCl₂ (oxidation of ethene to acetaldehyde). → II.
C. Wilkinson catalyst — [(PPh₃)₃RhCl] (homogeneous hydrogenation). → III.
D. Ziegler catalyst — TiCl₄ with Al(CH₃)₃ (polymerisation of olefins). → IV.

Hence the correct match is A-I, B-II, C-III, D-IV — option (3).