322251
The sum of number of unpaired electrons in a tetrahedral \({\mathrm{d^{6}}}\) ion and in a square planar \({\mathrm{\mathrm{d}^{7}}}\) ion are.
1 1
2 4
3 5
4 7
Explanation:
Electronic distribution in a tetrahedral \({\mathrm{d^{6}}}\) ion. Sum of unpaired electrons =4+1=5.
CHXII09:COORDINATION COMPOUNDS
322252
In hexacyanomanganate (II) ion the Mn-atom assumes \({{\text{d}}^{\text{2}}}{\text{s}}{{\text{p}}^{\text{3}}}\) - hybrid state. The number of unpaired electrons in the complex is:
1 2
2 3
3 0
4 1
Explanation:
\(\mathrm{Mn}\) in \(\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4}\) has configuration:
CHXII09:COORDINATION COMPOUNDS
322253
In which of the following complex ion, the central metal ion is in a state of \(\mathrm{sp}^{3} \mathrm{~d}^{2}\) hybridisation?
Electronic configuration of \(\mathrm{F}^{-}\), being a weak ligand cannot cause forcible pairing of electrons within \(d\)-subshell and forms outer orbital octahedral complex.
\(\mathrm{NH}_{3}\) and \(\mathrm{CN}^{-}\)are strong ligands, so they form inner orbital octahedral complex.
KCET - 2006
CHXII09:COORDINATION COMPOUNDS
322254
Assertion : Geometry of any complex depends upon the nature of ligands attached. Reason : \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) has square planar geometry while \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) has tetrahedral geometry.
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:
Electronic configuration of \(\mathrm{Ni}\) atom is \(3 d^{8} 4 s^{2}\) There is pairing of electrons in \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\). Thus \(\mathrm{Ni}(\mathrm{CO})_{4}\) is diamagnetic. It has \(\mathrm{sp}^{3}\) hybridization and tetrahedral geometry. In the complex, \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{-2}\) The electronic configuration of \(\mathrm{Ni}^{2+}\) is \(3 \mathrm{~d}^{8} 4 \mathrm{~s}^{0}\). Here pairing of electrons take place. Here pairing Thus \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) is diamagnetic with \(\mathrm{dsp}^{2}\) hybridization and square planar geometry So the option (3) is correct.
322251
The sum of number of unpaired electrons in a tetrahedral \({\mathrm{d^{6}}}\) ion and in a square planar \({\mathrm{\mathrm{d}^{7}}}\) ion are.
1 1
2 4
3 5
4 7
Explanation:
Electronic distribution in a tetrahedral \({\mathrm{d^{6}}}\) ion. Sum of unpaired electrons =4+1=5.
CHXII09:COORDINATION COMPOUNDS
322252
In hexacyanomanganate (II) ion the Mn-atom assumes \({{\text{d}}^{\text{2}}}{\text{s}}{{\text{p}}^{\text{3}}}\) - hybrid state. The number of unpaired electrons in the complex is:
1 2
2 3
3 0
4 1
Explanation:
\(\mathrm{Mn}\) in \(\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4}\) has configuration:
CHXII09:COORDINATION COMPOUNDS
322253
In which of the following complex ion, the central metal ion is in a state of \(\mathrm{sp}^{3} \mathrm{~d}^{2}\) hybridisation?
Electronic configuration of \(\mathrm{F}^{-}\), being a weak ligand cannot cause forcible pairing of electrons within \(d\)-subshell and forms outer orbital octahedral complex.
\(\mathrm{NH}_{3}\) and \(\mathrm{CN}^{-}\)are strong ligands, so they form inner orbital octahedral complex.
KCET - 2006
CHXII09:COORDINATION COMPOUNDS
322254
Assertion : Geometry of any complex depends upon the nature of ligands attached. Reason : \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) has square planar geometry while \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) has tetrahedral geometry.
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:
Electronic configuration of \(\mathrm{Ni}\) atom is \(3 d^{8} 4 s^{2}\) There is pairing of electrons in \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\). Thus \(\mathrm{Ni}(\mathrm{CO})_{4}\) is diamagnetic. It has \(\mathrm{sp}^{3}\) hybridization and tetrahedral geometry. In the complex, \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{-2}\) The electronic configuration of \(\mathrm{Ni}^{2+}\) is \(3 \mathrm{~d}^{8} 4 \mathrm{~s}^{0}\). Here pairing of electrons take place. Here pairing Thus \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) is diamagnetic with \(\mathrm{dsp}^{2}\) hybridization and square planar geometry So the option (3) is correct.
322251
The sum of number of unpaired electrons in a tetrahedral \({\mathrm{d^{6}}}\) ion and in a square planar \({\mathrm{\mathrm{d}^{7}}}\) ion are.
1 1
2 4
3 5
4 7
Explanation:
Electronic distribution in a tetrahedral \({\mathrm{d^{6}}}\) ion. Sum of unpaired electrons =4+1=5.
CHXII09:COORDINATION COMPOUNDS
322252
In hexacyanomanganate (II) ion the Mn-atom assumes \({{\text{d}}^{\text{2}}}{\text{s}}{{\text{p}}^{\text{3}}}\) - hybrid state. The number of unpaired electrons in the complex is:
1 2
2 3
3 0
4 1
Explanation:
\(\mathrm{Mn}\) in \(\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4}\) has configuration:
CHXII09:COORDINATION COMPOUNDS
322253
In which of the following complex ion, the central metal ion is in a state of \(\mathrm{sp}^{3} \mathrm{~d}^{2}\) hybridisation?
Electronic configuration of \(\mathrm{F}^{-}\), being a weak ligand cannot cause forcible pairing of electrons within \(d\)-subshell and forms outer orbital octahedral complex.
\(\mathrm{NH}_{3}\) and \(\mathrm{CN}^{-}\)are strong ligands, so they form inner orbital octahedral complex.
KCET - 2006
CHXII09:COORDINATION COMPOUNDS
322254
Assertion : Geometry of any complex depends upon the nature of ligands attached. Reason : \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) has square planar geometry while \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) has tetrahedral geometry.
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:
Electronic configuration of \(\mathrm{Ni}\) atom is \(3 d^{8} 4 s^{2}\) There is pairing of electrons in \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\). Thus \(\mathrm{Ni}(\mathrm{CO})_{4}\) is diamagnetic. It has \(\mathrm{sp}^{3}\) hybridization and tetrahedral geometry. In the complex, \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{-2}\) The electronic configuration of \(\mathrm{Ni}^{2+}\) is \(3 \mathrm{~d}^{8} 4 \mathrm{~s}^{0}\). Here pairing of electrons take place. Here pairing Thus \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) is diamagnetic with \(\mathrm{dsp}^{2}\) hybridization and square planar geometry So the option (3) is correct.
322251
The sum of number of unpaired electrons in a tetrahedral \({\mathrm{d^{6}}}\) ion and in a square planar \({\mathrm{\mathrm{d}^{7}}}\) ion are.
1 1
2 4
3 5
4 7
Explanation:
Electronic distribution in a tetrahedral \({\mathrm{d^{6}}}\) ion. Sum of unpaired electrons =4+1=5.
CHXII09:COORDINATION COMPOUNDS
322252
In hexacyanomanganate (II) ion the Mn-atom assumes \({{\text{d}}^{\text{2}}}{\text{s}}{{\text{p}}^{\text{3}}}\) - hybrid state. The number of unpaired electrons in the complex is:
1 2
2 3
3 0
4 1
Explanation:
\(\mathrm{Mn}\) in \(\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4}\) has configuration:
CHXII09:COORDINATION COMPOUNDS
322253
In which of the following complex ion, the central metal ion is in a state of \(\mathrm{sp}^{3} \mathrm{~d}^{2}\) hybridisation?
Electronic configuration of \(\mathrm{F}^{-}\), being a weak ligand cannot cause forcible pairing of electrons within \(d\)-subshell and forms outer orbital octahedral complex.
\(\mathrm{NH}_{3}\) and \(\mathrm{CN}^{-}\)are strong ligands, so they form inner orbital octahedral complex.
KCET - 2006
CHXII09:COORDINATION COMPOUNDS
322254
Assertion : Geometry of any complex depends upon the nature of ligands attached. Reason : \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) has square planar geometry while \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) has tetrahedral geometry.
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:
Electronic configuration of \(\mathrm{Ni}\) atom is \(3 d^{8} 4 s^{2}\) There is pairing of electrons in \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\). Thus \(\mathrm{Ni}(\mathrm{CO})_{4}\) is diamagnetic. It has \(\mathrm{sp}^{3}\) hybridization and tetrahedral geometry. In the complex, \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{-2}\) The electronic configuration of \(\mathrm{Ni}^{2+}\) is \(3 \mathrm{~d}^{8} 4 \mathrm{~s}^{0}\). Here pairing of electrons take place. Here pairing Thus \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) is diamagnetic with \(\mathrm{dsp}^{2}\) hybridization and square planar geometry So the option (3) is correct.
