MCQ
Elements in which $4f$ orbitals are progressively filled are called as
- ATransition elements
- ✓Lanthanides
- CActinides
- DInert gases
✓
Answer
Correct option: B.
Lanthanides
b
In lanthanoids, electrons are progressively filled in $4$ orbital. The general electronic configuration for lanthanoids is $[ Xe ] \,4 f ^{1-14} \,5 d ^{0-1}\, 6 s ^2$
In lanthanoids, electrons are progressively filled in $4$ orbital. The general electronic configuration for lanthanoids is $[ Xe ] \,4 f ^{1-14} \,5 d ^{0-1}\, 6 s ^2$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The configuration $1{s^2},\,2{s^2}2{p^5},\,3{s^1}$ shows
→Which of the following is not expected to be intermediate of the following reaction ?
Two mole of $HBr$ is added with $CH_3 -C$ $ \equiv $ $CH$ in presence of peroxide to give
→$\begin{matrix}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|| \\
HO-\underset{(3)}{\mathop{C{{H}_{2}}}}\,-\underset{(2)}{\mathop{C{{H}_{2}}}}\,-\underset{(1)}{\mathop{C}}\,-H \\
\end{matrix}$
→\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|| \\
HO-\underset{(3)}{\mathop{C{{H}_{2}}}}\,-\underset{(2)}{\mathop{C{{H}_{2}}}}\,-\underset{(1)}{\mathop{C}}\,-H \\
\end{matrix}$
Which conformer of above compound is most stable (consider conformer across $(C_2 -C_3)$
Chiral molecule is
Identify the product formed ( $A$ and $E$ )
→Which statement is correct in the following
In the following compound The favourable site/s for protonation is/are
Resistance of a conductivity cell (cell constant $129 \;m ^{-1}$ ) filled with $74.5\; ppm$ solution of $KCl$ is $100\; \Omega$ (labelled as solution 1). When the same cell is filled with $KCl$ solution of $149 \;ppm$, the resistance is $50\; \Omega$ (labelled as solution 2). The ratio of molar conductivity of solution $1$ and solution $2$ is i.e. $\frac{\wedge_{1}}{\wedge_{2}}=x \times 10^{-3}$. The value of $x$ is
→(Nearest integer)
Given, molar mass of $KCl$ is $74.5 g mol ^{-1}$
The potential energy of electron in third excited state of $He^+$ ion is ............. $\mathrm{eV}$
→