A - NaCl - B white ppt-B - KI - C green ppt-C - excessKI- D - E colourless solutionE - NH3 - KOH - FType of hybridization in compound E is-

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Sigma Bond and Pi Bond

A + NaCl → ...

Question

(A) + NaCl $\to$ (B) (white ppt.)

(B) + (KI) $\to$ (C) (green ppt.)

(C) + $K I (excess) \to$ (D) + (E) (colourless solution)

(E) + $N H_{3}$ + KOH $\to$ (F)

Type of hybridization in compound (E) is___________.

d^{2} s p^{3}

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s p^{3} d^{2}

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s p^{3}

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d s p^{2}

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Solution

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Similar questions

\to

\to

K I (excess) \to

N H_{3}

\to

A g N O_{3}

N H_{3}

H_{2} S

Z n C l_{2}

A g C l

Z n (O H)_{2}

N a_{2} Z n O_{2}

Z n S

N i C l_{2} + K C N (e x c e s s) \to A (c y a n o c o m p l e x)

N i C l_{2} + c o n c . H C l (e x c e s s) \to B (C h l o r o c o m p l e x)

\begin{matrix} Column-I & Column-II a. {[N i (C N)_{4}]}_{4}^{2 -} & p. d^{2} s p^{3} b. [N i (C O)_{4}] & q. s p^{3} d^{2} c. {[C o (e n)_{3}]}_{3}^{3 +} & r. d s p^{2} d. {[C o (B r)_{3} C l_{3}]}_{3}^{3 -} & s. s p^{3} t. Inner orbital complex u. Outer orbital complex \end{matrix}

M n O_{2}

K O H

K N O_{3}

K I

K I

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