The coordination number of $Ni^{2+}$ is $4$.
$NiCl_2 + KCN$ (excess) $\rightarrow A$ (cyano complex)
$NiCl_2 + \text{Conc. } HCl$ (excess) $\rightarrow B$ (chloro complex)
$1.$ The $IUPAC$ names of $A$ and $B$ are:
$(A)$ Potassium tetracyanonickelate $(II)$,potassium tetrachloronickelate $(II)$
$(B)$ Tetracyanopotassiumnickelate $(II)$,tetrachloropotassiumnickelate $(II)$
$(C)$ Tetracyanonickel $(II)$,tetrachloronickel $(II)$
$(D)$ Potassium tetracyanonickel $(II)$,potassium tetrachloronickel $(II)$
$2.$ Predict the magnetic nature of $A$ and $B$:
$(A)$ Both are diamagnetic.
$(B)$ $A$ is diamagnetic and $B$ is paramagnetic with one unpaired electron.
$(C)$ $A$ is diamagnetic and $B$ is paramagnetic with two unpaired electrons.
$(D)$ Both are paramagnetic.
$3.$ The hybridization of $A$ and $B$ are:
$(A)$ $dsp^2, sp^3$
$(B)$ $sp^3, sp^3$
$(C)$ $dsp^2, dsp^2$
$(D)$ $sp^3 d^2, d^2 sp^3$
Give the answers for questions $1, 2$ and $3$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(C) is $K_2[Ni(CN)_4]$ and $B$ is $K_2[NiCl_4]$.
$1.$ The $IUPAC$ name of $K_2[Ni(CN)_4]$ is Potassium tetracyanonickelate $(II)$ and $K_2[NiCl_4]$ is Potassium tetrachloronickelate $(II)$. Thus,option $(A)$ is correct.
$2.$ In $A$,$CN^-$ is a strong field ligand,so $Ni^{2+}$ $(3d^8)$ undergoes $dsp^2$ hybridization,making it diamagnetic. In $B$,$Cl^-$ is a weak field ligand,so $Ni^{2+}$ $(3d^8)$ undergoes $sp^3$ hybridization,leaving two unpaired electrons,making it paramagnetic. Thus,option $(C)$ is correct.
$3.$ As determined,$A$ has $dsp^2$ hybridization and $B$ has $sp^3$ hybridization. Thus,option $(A)$ is correct.
The sequence of answers is $A, C, A$.

Explore More

Similar Questions

Match List-$I$ with List-$II$ and select the correct option:
List-$I$List-$II$
$(A). [Ag(CN)_2]^-$$1. \text{Square planar, } 1.73 \, B.M.$
$(B). [Cu(CN)_4]^{3-}$$2. \text{Linear, } 0 \, B.M.$
$(C). [Cu(CN)_6]^{4-}$$3. \text{Octahedral, } 0 \, B.M.$
$(D). [Cu(NH_3)_4]^{2+}$$4. \text{Tetrahedral, } 0 \, B.M.$
$(E). [Fe(CN)_6]^{4-}$$5. \text{Octahedral, } 1.73 \, B.M.$

Difficult
View Solution

Which of the following has a yellow colour?

The colour of $CuCr_2O_7$ solution in water is green because

Among the statements $(a)-(d)$,the incorrect ones are:
$(a)$ Octahedral $Co(III)$ complexes with strong field ligands have very high magnetic moments.
$(b)$ When $\Delta_{0} < P$,the $d-$electron configuration of $Co(III)$ in an octahedral complex is $t_{2g}^{4} e_{g}^{2}$.
$(c)$ Wavelength of light absorbed by $[Co(en)_{3}]^{3+}$ is lower than that of $[CoF_{6}]^{3-}$.
$(d)$ If the $\Delta_{0}$ for an octahedral complex of $Co(III)$ is $18,000 \ cm^{-1}$,the $\Delta_{t}$ for its tetrahedral complex with the same ligand will be $16,000 \ cm^{-1}$.

When a metal rod $M$ is dipped into an aqueous colourless concentrated solution of compound $N$,the solution turns light blue. Addition of aqueous $NaCl$ to the blue solution gives a white precipitate $O$. Addition of aqueous $NH_3$ dissolves $O$ and gives an intense blue solution.
$1.$ The metal rod $M$ is
$(A)$ $Fe$ $(B)$ $Cu$ $(C)$ $Ni$ $(D)$ $Co$
$2.$ The compound $N$ is
$(A)$ $AgNO_3$ $(B)$ $Zn(NO_3)_2$
$(C)$ $Al(NO_3)_3$ $(D)$ $Pb(NO_3)_2$
$3.$ The final solution contains
$(A)$ $[Pb(NH_3)_4]^{2+}$ and $[CoCl_4]^{2-}$
$(B)$ $[Al(NH_3)_4]^{3+}$ and $[Cu(NH_3)_4]^{2+}$
$(C)$ $[Ag(NH_3)_2]^{+}$ and $[Cu(NH_3)_4]^{2+}$
$(D)$ $[Ag(NH_3)_2]^{+}$ and $[Ni(NH_3)_6]^{2+}$
Give the answer for questions $1$,$2$ and $3$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo