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List-I having complex compounds, List-II having hybridization / no. of unpaired electrons.
LIST-I
LIST-II
I)
$N i {(C O)}_{4}$
P)
$s p^{3} d^{2}$
II)
${[N i F_{6}]}^{- 2}$
Q)
$d^{2} s p^{3}$
III)
${[F e F_{6}]}^{- 4}$
R)
Even no. of unpaired electrons.
IV)
${[C o {(H_{2} O)}_{6}]}^{+ 2}$
S)
Odd no. unpaired electrons.

T)
$s p^{3}$

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I–T, II–Q, III–P, IV–S

I–T, II–Q, III–R, IV–Q

I–T, II–P, III–R, IV–P

I–Q, II–T, III–P, IV–P

answer is A.

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Detailed Solution

${[C o {(H_{2} O)}_{6}]}^{+ 2}$ is $s p^{3} d^{2}$

Nickel in +4 oxidation will pair up electrons even in the presence of fluoride, which is exception.

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