322225
Assertion : All the octahedral complexes of \(\mathrm{Ni}^{2+}\) must be outer orbital complexes. Reason : Outer orbital octahedral complexes are given by weak ligands.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
\(\mathrm{Ni}\) has +2 state in its octahedral complexes. Electronic configuration of \(\mathrm{Ni}^{+2}\) is \([\mathrm{Ar}] 3 \mathrm{~d}^{8}\) During rearrangement only one \(\mathrm{d}\) orbital may be made available by pairing the electrons. Thus, inner hybridization of \(\mathrm{d}^{2} \mathrm{sp}^{3}\) is not possible. Thus, only outer hybridization of \(\mathrm{sp}^{3} \mathrm{~d}^{2}\) can occur. So the option (2) is correct.
CHXII09:COORDINATION COMPOUNDS
322226
Assertion : \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) is square planar. Reason : The oxidation state of platinum is +2 .
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
In \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) the central metal, \(\mathrm{Pt}\) is in the +2 state. It undergoes \(\mathrm{dsp}^{2}\) hybridization in order to accommodate the electron pairs from the four ligands. The two strong ammonia ligands force pairing of electrons and creates empty spaces in the inner subshell. So the option (2) is correct.
CHXII09:COORDINATION COMPOUNDS
322227
Pick out the correct statement with respect to \(\left.\left[\mathrm{Mn}(\mathrm{CN})_{6}\right)\right]^{3-}:-\)
1 It is \({\text{s}}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}\) hybridised and tetrahedral
2 It is \({{\text{d}}^{\text{2}}}{\text{s}}{{\text{p}}^{\text{3}}}\) hybridised and octahedral
3 It is \({\text{ds}}{{\text{p}}^{\text{2}}}\) hybridised and square planar
4 It is \({\text{s}}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}\) hybridised and octahedral
Explanation:
\(\left.\left[\mathrm{Mn}(\mathrm{CN})_{6}\right)\right]^{3-} \rightarrow O . S\). of \(\mathrm{Mn}\) is (+3) C.N. \(=6\)
Presence of SFL (Pairing is possible) octahedral)
NEET - 2017
CHXII09:COORDINATION COMPOUNDS
322228
The geometry of \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) and \(\left[\mathrm{Ni}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) are
1 both square planar
2 tetrahedral and square planar respectively
3 both tetrahedral
4 square planar and tetrahedral respectively
Explanation:
\(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) and \(\left[\mathrm{Ni}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) are tetrahedral in geometrical shape, because coordination number of \(\mathrm{Ni}\) is 4 in both cases and both use 3s and 3p-orbitals of the central metal to give \({\rm{s}}{{\rm{p}}^{\rm{3}}}\)-hybridisation.
CHXII09:COORDINATION COMPOUNDS
322229
Both \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) and \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) are diamagnetic. The hybridisation of nickel in these complex, respectively are
In carbonyls O.S. of metal is zero In \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\), the oxidation state of nickel is zero. Its configuration in \(\mathrm{Ni}(\mathrm{CO})_{4}\) is In \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) the oxidation state of \(\mathrm{Ni}\) is +2 and its configuration is Thus the hybridisation of nickel in these compounds are \({\text{s}}{{\text{p}}^{\text{3}}}\,\,{\text{and}}\,\,{\text{ds}}{{\text{p}}^{\text{2}}}\) respectively.
322225
Assertion : All the octahedral complexes of \(\mathrm{Ni}^{2+}\) must be outer orbital complexes. Reason : Outer orbital octahedral complexes are given by weak ligands.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
\(\mathrm{Ni}\) has +2 state in its octahedral complexes. Electronic configuration of \(\mathrm{Ni}^{+2}\) is \([\mathrm{Ar}] 3 \mathrm{~d}^{8}\) During rearrangement only one \(\mathrm{d}\) orbital may be made available by pairing the electrons. Thus, inner hybridization of \(\mathrm{d}^{2} \mathrm{sp}^{3}\) is not possible. Thus, only outer hybridization of \(\mathrm{sp}^{3} \mathrm{~d}^{2}\) can occur. So the option (2) is correct.
CHXII09:COORDINATION COMPOUNDS
322226
Assertion : \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) is square planar. Reason : The oxidation state of platinum is +2 .
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
In \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) the central metal, \(\mathrm{Pt}\) is in the +2 state. It undergoes \(\mathrm{dsp}^{2}\) hybridization in order to accommodate the electron pairs from the four ligands. The two strong ammonia ligands force pairing of electrons and creates empty spaces in the inner subshell. So the option (2) is correct.
CHXII09:COORDINATION COMPOUNDS
322227
Pick out the correct statement with respect to \(\left.\left[\mathrm{Mn}(\mathrm{CN})_{6}\right)\right]^{3-}:-\)
1 It is \({\text{s}}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}\) hybridised and tetrahedral
2 It is \({{\text{d}}^{\text{2}}}{\text{s}}{{\text{p}}^{\text{3}}}\) hybridised and octahedral
3 It is \({\text{ds}}{{\text{p}}^{\text{2}}}\) hybridised and square planar
4 It is \({\text{s}}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}\) hybridised and octahedral
Explanation:
\(\left.\left[\mathrm{Mn}(\mathrm{CN})_{6}\right)\right]^{3-} \rightarrow O . S\). of \(\mathrm{Mn}\) is (+3) C.N. \(=6\)
Presence of SFL (Pairing is possible) octahedral)
NEET - 2017
CHXII09:COORDINATION COMPOUNDS
322228
The geometry of \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) and \(\left[\mathrm{Ni}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) are
1 both square planar
2 tetrahedral and square planar respectively
3 both tetrahedral
4 square planar and tetrahedral respectively
Explanation:
\(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) and \(\left[\mathrm{Ni}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) are tetrahedral in geometrical shape, because coordination number of \(\mathrm{Ni}\) is 4 in both cases and both use 3s and 3p-orbitals of the central metal to give \({\rm{s}}{{\rm{p}}^{\rm{3}}}\)-hybridisation.
CHXII09:COORDINATION COMPOUNDS
322229
Both \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) and \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) are diamagnetic. The hybridisation of nickel in these complex, respectively are
In carbonyls O.S. of metal is zero In \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\), the oxidation state of nickel is zero. Its configuration in \(\mathrm{Ni}(\mathrm{CO})_{4}\) is In \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) the oxidation state of \(\mathrm{Ni}\) is +2 and its configuration is Thus the hybridisation of nickel in these compounds are \({\text{s}}{{\text{p}}^{\text{3}}}\,\,{\text{and}}\,\,{\text{ds}}{{\text{p}}^{\text{2}}}\) respectively.
322225
Assertion : All the octahedral complexes of \(\mathrm{Ni}^{2+}\) must be outer orbital complexes. Reason : Outer orbital octahedral complexes are given by weak ligands.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
\(\mathrm{Ni}\) has +2 state in its octahedral complexes. Electronic configuration of \(\mathrm{Ni}^{+2}\) is \([\mathrm{Ar}] 3 \mathrm{~d}^{8}\) During rearrangement only one \(\mathrm{d}\) orbital may be made available by pairing the electrons. Thus, inner hybridization of \(\mathrm{d}^{2} \mathrm{sp}^{3}\) is not possible. Thus, only outer hybridization of \(\mathrm{sp}^{3} \mathrm{~d}^{2}\) can occur. So the option (2) is correct.
CHXII09:COORDINATION COMPOUNDS
322226
Assertion : \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) is square planar. Reason : The oxidation state of platinum is +2 .
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
In \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) the central metal, \(\mathrm{Pt}\) is in the +2 state. It undergoes \(\mathrm{dsp}^{2}\) hybridization in order to accommodate the electron pairs from the four ligands. The two strong ammonia ligands force pairing of electrons and creates empty spaces in the inner subshell. So the option (2) is correct.
CHXII09:COORDINATION COMPOUNDS
322227
Pick out the correct statement with respect to \(\left.\left[\mathrm{Mn}(\mathrm{CN})_{6}\right)\right]^{3-}:-\)
1 It is \({\text{s}}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}\) hybridised and tetrahedral
2 It is \({{\text{d}}^{\text{2}}}{\text{s}}{{\text{p}}^{\text{3}}}\) hybridised and octahedral
3 It is \({\text{ds}}{{\text{p}}^{\text{2}}}\) hybridised and square planar
4 It is \({\text{s}}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}\) hybridised and octahedral
Explanation:
\(\left.\left[\mathrm{Mn}(\mathrm{CN})_{6}\right)\right]^{3-} \rightarrow O . S\). of \(\mathrm{Mn}\) is (+3) C.N. \(=6\)
Presence of SFL (Pairing is possible) octahedral)
NEET - 2017
CHXII09:COORDINATION COMPOUNDS
322228
The geometry of \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) and \(\left[\mathrm{Ni}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) are
1 both square planar
2 tetrahedral and square planar respectively
3 both tetrahedral
4 square planar and tetrahedral respectively
Explanation:
\(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) and \(\left[\mathrm{Ni}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) are tetrahedral in geometrical shape, because coordination number of \(\mathrm{Ni}\) is 4 in both cases and both use 3s and 3p-orbitals of the central metal to give \({\rm{s}}{{\rm{p}}^{\rm{3}}}\)-hybridisation.
CHXII09:COORDINATION COMPOUNDS
322229
Both \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) and \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) are diamagnetic. The hybridisation of nickel in these complex, respectively are
In carbonyls O.S. of metal is zero In \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\), the oxidation state of nickel is zero. Its configuration in \(\mathrm{Ni}(\mathrm{CO})_{4}\) is In \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) the oxidation state of \(\mathrm{Ni}\) is +2 and its configuration is Thus the hybridisation of nickel in these compounds are \({\text{s}}{{\text{p}}^{\text{3}}}\,\,{\text{and}}\,\,{\text{ds}}{{\text{p}}^{\text{2}}}\) respectively.
NEET Test Series from KOTA - 10 Papers In MS WORD
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CHXII09:COORDINATION COMPOUNDS
322225
Assertion : All the octahedral complexes of \(\mathrm{Ni}^{2+}\) must be outer orbital complexes. Reason : Outer orbital octahedral complexes are given by weak ligands.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
\(\mathrm{Ni}\) has +2 state in its octahedral complexes. Electronic configuration of \(\mathrm{Ni}^{+2}\) is \([\mathrm{Ar}] 3 \mathrm{~d}^{8}\) During rearrangement only one \(\mathrm{d}\) orbital may be made available by pairing the electrons. Thus, inner hybridization of \(\mathrm{d}^{2} \mathrm{sp}^{3}\) is not possible. Thus, only outer hybridization of \(\mathrm{sp}^{3} \mathrm{~d}^{2}\) can occur. So the option (2) is correct.
CHXII09:COORDINATION COMPOUNDS
322226
Assertion : \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) is square planar. Reason : The oxidation state of platinum is +2 .
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
In \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) the central metal, \(\mathrm{Pt}\) is in the +2 state. It undergoes \(\mathrm{dsp}^{2}\) hybridization in order to accommodate the electron pairs from the four ligands. The two strong ammonia ligands force pairing of electrons and creates empty spaces in the inner subshell. So the option (2) is correct.
CHXII09:COORDINATION COMPOUNDS
322227
Pick out the correct statement with respect to \(\left.\left[\mathrm{Mn}(\mathrm{CN})_{6}\right)\right]^{3-}:-\)
1 It is \({\text{s}}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}\) hybridised and tetrahedral
2 It is \({{\text{d}}^{\text{2}}}{\text{s}}{{\text{p}}^{\text{3}}}\) hybridised and octahedral
3 It is \({\text{ds}}{{\text{p}}^{\text{2}}}\) hybridised and square planar
4 It is \({\text{s}}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}\) hybridised and octahedral
Explanation:
\(\left.\left[\mathrm{Mn}(\mathrm{CN})_{6}\right)\right]^{3-} \rightarrow O . S\). of \(\mathrm{Mn}\) is (+3) C.N. \(=6\)
Presence of SFL (Pairing is possible) octahedral)
NEET - 2017
CHXII09:COORDINATION COMPOUNDS
322228
The geometry of \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) and \(\left[\mathrm{Ni}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) are
1 both square planar
2 tetrahedral and square planar respectively
3 both tetrahedral
4 square planar and tetrahedral respectively
Explanation:
\(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) and \(\left[\mathrm{Ni}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) are tetrahedral in geometrical shape, because coordination number of \(\mathrm{Ni}\) is 4 in both cases and both use 3s and 3p-orbitals of the central metal to give \({\rm{s}}{{\rm{p}}^{\rm{3}}}\)-hybridisation.
CHXII09:COORDINATION COMPOUNDS
322229
Both \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) and \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) are diamagnetic. The hybridisation of nickel in these complex, respectively are
In carbonyls O.S. of metal is zero In \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\), the oxidation state of nickel is zero. Its configuration in \(\mathrm{Ni}(\mathrm{CO})_{4}\) is In \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) the oxidation state of \(\mathrm{Ni}\) is +2 and its configuration is Thus the hybridisation of nickel in these compounds are \({\text{s}}{{\text{p}}^{\text{3}}}\,\,{\text{and}}\,\,{\text{ds}}{{\text{p}}^{\text{2}}}\) respectively.
322225
Assertion : All the octahedral complexes of \(\mathrm{Ni}^{2+}\) must be outer orbital complexes. Reason : Outer orbital octahedral complexes are given by weak ligands.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
\(\mathrm{Ni}\) has +2 state in its octahedral complexes. Electronic configuration of \(\mathrm{Ni}^{+2}\) is \([\mathrm{Ar}] 3 \mathrm{~d}^{8}\) During rearrangement only one \(\mathrm{d}\) orbital may be made available by pairing the electrons. Thus, inner hybridization of \(\mathrm{d}^{2} \mathrm{sp}^{3}\) is not possible. Thus, only outer hybridization of \(\mathrm{sp}^{3} \mathrm{~d}^{2}\) can occur. So the option (2) is correct.
CHXII09:COORDINATION COMPOUNDS
322226
Assertion : \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) is square planar. Reason : The oxidation state of platinum is +2 .
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
In \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) the central metal, \(\mathrm{Pt}\) is in the +2 state. It undergoes \(\mathrm{dsp}^{2}\) hybridization in order to accommodate the electron pairs from the four ligands. The two strong ammonia ligands force pairing of electrons and creates empty spaces in the inner subshell. So the option (2) is correct.
CHXII09:COORDINATION COMPOUNDS
322227
Pick out the correct statement with respect to \(\left.\left[\mathrm{Mn}(\mathrm{CN})_{6}\right)\right]^{3-}:-\)
1 It is \({\text{s}}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}\) hybridised and tetrahedral
2 It is \({{\text{d}}^{\text{2}}}{\text{s}}{{\text{p}}^{\text{3}}}\) hybridised and octahedral
3 It is \({\text{ds}}{{\text{p}}^{\text{2}}}\) hybridised and square planar
4 It is \({\text{s}}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}\) hybridised and octahedral
Explanation:
\(\left.\left[\mathrm{Mn}(\mathrm{CN})_{6}\right)\right]^{3-} \rightarrow O . S\). of \(\mathrm{Mn}\) is (+3) C.N. \(=6\)
Presence of SFL (Pairing is possible) octahedral)
NEET - 2017
CHXII09:COORDINATION COMPOUNDS
322228
The geometry of \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) and \(\left[\mathrm{Ni}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) are
1 both square planar
2 tetrahedral and square planar respectively
3 both tetrahedral
4 square planar and tetrahedral respectively
Explanation:
\(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) and \(\left[\mathrm{Ni}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) are tetrahedral in geometrical shape, because coordination number of \(\mathrm{Ni}\) is 4 in both cases and both use 3s and 3p-orbitals of the central metal to give \({\rm{s}}{{\rm{p}}^{\rm{3}}}\)-hybridisation.
CHXII09:COORDINATION COMPOUNDS
322229
Both \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) and \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) are diamagnetic. The hybridisation of nickel in these complex, respectively are
In carbonyls O.S. of metal is zero In \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\), the oxidation state of nickel is zero. Its configuration in \(\mathrm{Ni}(\mathrm{CO})_{4}\) is In \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) the oxidation state of \(\mathrm{Ni}\) is +2 and its configuration is Thus the hybridisation of nickel in these compounds are \({\text{s}}{{\text{p}}^{\text{3}}}\,\,{\text{and}}\,\,{\text{ds}}{{\text{p}}^{\text{2}}}\) respectively.
