NEET
Practice Test 12
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Which one of the following is an outer orbital complex and exhibits paramagnetic behaviour ?

  • Solution

    [Ni(NH3)6]2+

    Ni2+ = 3d8, according to CFT = \(t_{2g}^{6}e_{g}^{2}\) therefore,hybridisation is sp3d2 and complex is paramagnetic.

Which of the following carbonyls will have the strongest C – O bond ?

  • Solution

    As positive charge on the central metal atom increases,the less readily the metal can donate electron density into the π* orbitals of CO ligand (donation of electron density into π* orbitals of CO result in weakening of C – O bond). Hence, the C – O bond would be strongest in [Mn(CO)6]+.

The d-electron configurations of Cr2+, Mn2+, Fe2+ and Co2+ are d4, d5, d6 and d7, respectively. Which one of the following will exhibit minimum paramagnetic behaviour?

(At,nos.Cr = 24, Mn = 25, Fe = 26,Co = 27)

  • Solution

The existence of two different coloured complexes with the composition of [Co(NH3)4Cl2]+ is due to :

  • Solution

Crystalfield stabilization energy for high spin d4 octahedral complex is:

  • Solution

    Where, x → electrons in t2g orbital
    y → electrons in eg orbital

    CFSE = [0.6 × 1] + [–0.4 × 3] = – 0.6 Δ0

0.02 mole of[Co(NH3)5Br]Cl2 and 0.02 mole of [Co(NH3)5Cl]SO4 are present in 200 cc of a solution X. The number of moles of the precipitates Y and Z that are formed when the solution X is treated with excess silver nitrate and excess barium chloride are respectively

  • Solution

Among the following, the species that is both paramagnetic and coloured is –

  • Solution

[Fe(en)2(H2O)2]2+ + en → complex (X). The correct statement about the complex (X) is –

  • Solution

A complex cation is formed by Pt (in some oxidation state)with ligands (in proper number so that coordination number of Pt becomes six). Which of the following can be its correct IUPAC name ?

  • Solution

    (1)[Pt(en) (SCN)2(NH3)2]0; it is not ion

    (2)[Pt(en) (SCN)2(NH3)2]2+; it should not be platinate

    (3)[Pt (en) (SCN)2(NH3)2]2+: correct name

    (4)[Pt(en)2(SCN)2(NH3)2]2+; its coordination number is 8

Nickel (Z = 28) combines with a uninegative monodentatelig and X–to form a paramagnetic complex[NiX4]2–. The number of unpaired electron(s) in the nickel and geometry of this complex ion are, respectively

  • Solution

    [Ni X4]2–, the electronic configuration of Ni2+ is

    It contains two unpaired electrons and the hybridisation is sp3

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