The number of isoelectronic species among Sc^ | vedclass

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(B) SpeciesElectrons$Sc^{3+}$$18$$Cr^{2+}$$22$$Mn^{3+}$$22$$Co^{3+}$$24$$Fe^{3+}$$23$The isoelectronic species are $Cr^{2+}$ and $Mn^{3+}$, so $n = 2$

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) SpeciesElectrons$Sc^{3+}$$18$$Cr^{2+}$$22$$Mn^{3+}$$22$$Co^{3+}$$24$$Fe^{3+}$$23$The isoelectronic species are $Cr^{2+}$ and $Mn^{3+}$, so $n = 2$.The complex $CoCl_{3}(en)_{2}NH_{3}$ reacts with excess $AgNO_{3}$ to form $2$ moles of $AgCl$, indicating the formula is $[Co(en)_{2}NH_{3}Cl]Cl_{2}$.In this complex, $Co$ is in the $+3$ oxidation state ($3d^{6}$ configuration).For an octahedral complex with a strong field ligand like $en$, the $6$ electrons occupy the $t_{2g}$ orbitals as $t_{2g}^{6} e_{g}^{0}$.Therefore, the number of electrons in the $t_{2g}$ orbital is $6$.

Coordination number and type of hybridisation	Distribution of hybrid orbitals in space
$(a)$. $4, sp^3$	$(i)$. trigonal bipyramidal
$(b)$. $3, sp^2$	$(ii)$. octahedral
$(c)$. $5, sp^3 d$	$(iii)$. tetrahedral
$(d)$. $6, sp^3 d^2$	$(iv)$. trigonal planar

Similar Questions

Which of the following complexes has a square planar geometry and $dsp^2$ hybridization?

The number of species from the following that are involved in $sp^3 d^2$ hybridization is $[Co(NH_3)_6]^{3+}, SF_6, [CrF_6]^{3-}, [CoF_6]^{3-}, [Mn(CN)_6]^{3-}$ and $[MnCl_6]^{3-}$.

Which of the following has a square planar geometry?
(At. nos.: $Fe=26, Co=27, Ni=28, Pt=78$)

Statement $1$: $[Ti(H_2O)_6]^{4+}$ is coloured while $[Sc(H_2O)_6]^{3+}$ is colourless.
Statement $2$: $d-d$ transition is not possible in $[Sc(H_2O)_6]^{3+}$.

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