Statement 1: [Ti(H_2O)_6]^{4+} is coloured wh | vedclass

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(D) In $[Ti(H_2O)_6]^{4+}$,the oxidation state of $Ti$ is $+4$. The electronic configuration of $Ti^{4+}$ is $[Ar]3d^0$. Since there are no $d$-electr

Vedclass · Accepted Answer

Statement-$1$ is False,Statement-$2$ is True.

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(D) In $[Ti(H_2O)_6]^{4+}$,the oxidation state of $Ti$ is $+4$. The electronic configuration of $Ti^{4+}$ is $[Ar]3d^0$. Since there are no $d$-electrons,$d-d$ transition is not possible,and it is colourless. Thus,Statement-$1$ is False.In $[Sc(H_2O)_6]^{3+}$,the oxidation state of $Sc$ is $+3$. The electronic configuration of $Sc^{3+}$ is $[Ar]3d^0$. Since there are no $d$-electrons,$d-d$ transition is not possible,and it is colourless. Thus,Statement-$2$ is True.Therefore,Statement-$1$ is False and Statement-$2$ is True.

List-$I$ (Complex)	List-$II$ (Hybridization)
$A. Ni(CO)_4$	$I. sp^3d^2$
$B. [Ni(CN)_4]^{2-}$	$II. d^2sp^3$
$C. [Co(NH_3)_6]^{3+}$	$III. dsp^2$
$D. [CoF_6]^{3-}$	$IV. sp^3$

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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An octahedral complex of $Ni(II)$ must be:

Match the complexes in List-$I$ with their hybridization in List-$II$.
List-$I$ (Complex) List-$II$ (Hybridization)
$A. Ni(CO)_4$ $I. sp^3d^2$
$B. [Ni(CN)_4]^{2-}$ $II. d^2sp^3$
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Which of the following pairs of structure and magnetic properties is correct for the tetracyanonickelate$(II)$ ion?

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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 $[Sc(H_2O)_6]^{3+}$.