| A.\[Ce{{(OH)}_{3}}\] |
| B.\[Dy{{(OH)}_{3}}\] |
| C \[\Pr {{(OH)}_{3}}\] |
| D.\[Lu{{(OH)}_{3}}\] |
A) A > B > C > D
B) A > C > B > D
C) D > C > B > A
D) C > A > B > D
Correct Answer: B
Solution :
Charge density increases from \[=(\ell n\,a)\,\underset{x\to 0}{\mathop{Lim}}\,\,\frac{1}{\frac{(1-\cos x)}{x}+\frac{x}{x}}\,=\ell n\,a\] to \[=(\ell n\,a)\,\underset{x\to 0}{\mathop{Lim}}\,\,\frac{x-1}{\ell n(x)}\] \[x=1+h\] Ce-0 bond becomes stronger hence becomes weaker in basic nature.
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