KVPY SX SB Exam 2015 Question Paper

Question : 103 of 120

Marks: +1, -0

The geometry and the number of unpaired electron(s) of

[MnBr_4]^{2–}

, respectively, are

Solution:

[MnBr_4]^{2–}

x + (–1)4 = –2
x = +2

Mn\ {–––––▸}4s^23d^5

Mn^{+2}\ {–––––▸} 4s°3d^5

We know that Br is weak field ligand so Hund's Rule is applicable

Hybridization is

\ {–––––▸} sp^3

and N

_0

of unpaired = 5 electron.

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