28739
What type of hybridization is involved in \({[Fe{(CN)_6}]^{3 - }}\)
1 \({d^2}s{p^3}\)
2 \(ds{p^2}\)
3 \(s{p^3}{d^2}\)
4 \(ds{p^3}\)
Explanation:
It’s obvious.
COORDINATION COMPOUNDS
28740
The example of \(ds{p^2}\) hybridisation is
1 \(Fe(CN)_6^{3 - }\)
2 \(Ni\left( {CN} \right)_4^{2 - }\)
3 \(Zn\left( {N{H_3}} \right)_4^{2 + }\)
4 \(FeF_6^{3 - }\)
Explanation:
\(\left[ Ni ( CN )_4\right]^{2-}\) exhibit \(d s^2\) geometry and hence sqaure planar geometry. Due to strong field ligand \(CN ^{-}\)electrons get paired in \(3 d\) orbitals.
COORDINATION COMPOUNDS
28741
The shape of \({\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }}\) is square planar, \(C{u^{2 + }}\) in this complex is
1 \(s{p^3}\) hybridised
2 \(ds{p^2}\) hybridised
3 \(s{p^3}d\) hybridised
4 \(s{p^3}{d^2}\) hybridised
Explanation:
(b)Due to formation of inner orbital complex.
COORDINATION COMPOUNDS
28742
The geometry of \(Ni{\left( {CO} \right)_4}\) and \(Ni{\left( {PP{h_3}} \right)_2}C{l_2}\) are
1 Both square planar
2 Tetrahedral and square planar respectively
3 Square planar and tetrahedral respectively
4 Both tetrahedral
Explanation:
\(N i(C O)_{4}\) and \(N i\left(P P h_{3}\right)_{2} C l_{2}\) are tetrahedral in geometrical shape, because co-ordination number of \(Ni\) is 4 in both cases.
28739
What type of hybridization is involved in \({[Fe{(CN)_6}]^{3 - }}\)
1 \({d^2}s{p^3}\)
2 \(ds{p^2}\)
3 \(s{p^3}{d^2}\)
4 \(ds{p^3}\)
Explanation:
It’s obvious.
COORDINATION COMPOUNDS
28740
The example of \(ds{p^2}\) hybridisation is
1 \(Fe(CN)_6^{3 - }\)
2 \(Ni\left( {CN} \right)_4^{2 - }\)
3 \(Zn\left( {N{H_3}} \right)_4^{2 + }\)
4 \(FeF_6^{3 - }\)
Explanation:
\(\left[ Ni ( CN )_4\right]^{2-}\) exhibit \(d s^2\) geometry and hence sqaure planar geometry. Due to strong field ligand \(CN ^{-}\)electrons get paired in \(3 d\) orbitals.
COORDINATION COMPOUNDS
28741
The shape of \({\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }}\) is square planar, \(C{u^{2 + }}\) in this complex is
1 \(s{p^3}\) hybridised
2 \(ds{p^2}\) hybridised
3 \(s{p^3}d\) hybridised
4 \(s{p^3}{d^2}\) hybridised
Explanation:
(b)Due to formation of inner orbital complex.
COORDINATION COMPOUNDS
28742
The geometry of \(Ni{\left( {CO} \right)_4}\) and \(Ni{\left( {PP{h_3}} \right)_2}C{l_2}\) are
1 Both square planar
2 Tetrahedral and square planar respectively
3 Square planar and tetrahedral respectively
4 Both tetrahedral
Explanation:
\(N i(C O)_{4}\) and \(N i\left(P P h_{3}\right)_{2} C l_{2}\) are tetrahedral in geometrical shape, because co-ordination number of \(Ni\) is 4 in both cases.
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COORDINATION COMPOUNDS
28739
What type of hybridization is involved in \({[Fe{(CN)_6}]^{3 - }}\)
1 \({d^2}s{p^3}\)
2 \(ds{p^2}\)
3 \(s{p^3}{d^2}\)
4 \(ds{p^3}\)
Explanation:
It’s obvious.
COORDINATION COMPOUNDS
28740
The example of \(ds{p^2}\) hybridisation is
1 \(Fe(CN)_6^{3 - }\)
2 \(Ni\left( {CN} \right)_4^{2 - }\)
3 \(Zn\left( {N{H_3}} \right)_4^{2 + }\)
4 \(FeF_6^{3 - }\)
Explanation:
\(\left[ Ni ( CN )_4\right]^{2-}\) exhibit \(d s^2\) geometry and hence sqaure planar geometry. Due to strong field ligand \(CN ^{-}\)electrons get paired in \(3 d\) orbitals.
COORDINATION COMPOUNDS
28741
The shape of \({\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }}\) is square planar, \(C{u^{2 + }}\) in this complex is
1 \(s{p^3}\) hybridised
2 \(ds{p^2}\) hybridised
3 \(s{p^3}d\) hybridised
4 \(s{p^3}{d^2}\) hybridised
Explanation:
(b)Due to formation of inner orbital complex.
COORDINATION COMPOUNDS
28742
The geometry of \(Ni{\left( {CO} \right)_4}\) and \(Ni{\left( {PP{h_3}} \right)_2}C{l_2}\) are
1 Both square planar
2 Tetrahedral and square planar respectively
3 Square planar and tetrahedral respectively
4 Both tetrahedral
Explanation:
\(N i(C O)_{4}\) and \(N i\left(P P h_{3}\right)_{2} C l_{2}\) are tetrahedral in geometrical shape, because co-ordination number of \(Ni\) is 4 in both cases.
28739
What type of hybridization is involved in \({[Fe{(CN)_6}]^{3 - }}\)
1 \({d^2}s{p^3}\)
2 \(ds{p^2}\)
3 \(s{p^3}{d^2}\)
4 \(ds{p^3}\)
Explanation:
It’s obvious.
COORDINATION COMPOUNDS
28740
The example of \(ds{p^2}\) hybridisation is
1 \(Fe(CN)_6^{3 - }\)
2 \(Ni\left( {CN} \right)_4^{2 - }\)
3 \(Zn\left( {N{H_3}} \right)_4^{2 + }\)
4 \(FeF_6^{3 - }\)
Explanation:
\(\left[ Ni ( CN )_4\right]^{2-}\) exhibit \(d s^2\) geometry and hence sqaure planar geometry. Due to strong field ligand \(CN ^{-}\)electrons get paired in \(3 d\) orbitals.
COORDINATION COMPOUNDS
28741
The shape of \({\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }}\) is square planar, \(C{u^{2 + }}\) in this complex is
1 \(s{p^3}\) hybridised
2 \(ds{p^2}\) hybridised
3 \(s{p^3}d\) hybridised
4 \(s{p^3}{d^2}\) hybridised
Explanation:
(b)Due to formation of inner orbital complex.
COORDINATION COMPOUNDS
28742
The geometry of \(Ni{\left( {CO} \right)_4}\) and \(Ni{\left( {PP{h_3}} \right)_2}C{l_2}\) are
1 Both square planar
2 Tetrahedral and square planar respectively
3 Square planar and tetrahedral respectively
4 Both tetrahedral
Explanation:
\(N i(C O)_{4}\) and \(N i\left(P P h_{3}\right)_{2} C l_{2}\) are tetrahedral in geometrical shape, because co-ordination number of \(Ni\) is 4 in both cases.