Question

Which of the following are correct?
(1) Electron density in $X Y$ plane for $d_{x^{2}-y^{2}}$ orbital is zero.
(2) The energy of $3 p$ - orbital is higher than the energy of $2 p$ - orbital.
(3) $3 p_{z}$ -orbital has one angular node.
(4) $4 f$ -orbital has no radial node.

• A. 1,2,3,4
• B. 2,3,1
• C. 2,3,4
• D. 3,4,1

Answer 1

Answer: (C)

1. In $d_{x^{2}-y^{2}}$ orbital, electrons are present on the $x$ and $y$ axis, hence electron density is not zero.

2. According to $(n +l)$ rule, higher be the sum of $(n +l)$, higher is the energy.

$\because$ For $3 p,(n +l)=3+1=4$

and for $2 p,(n +l)=2+1=3$

Thus, $3 p$ has higher energy than $2 p$.

3. The quantum number $l$ determines the number of angular nodes. (It is also known as nodal plane)

i.e. Number of angular nodes $=$ value of $l$.

$\because$ For $3 p_{z}$ and $l=1$,

Thus, one angular node is present in $3 p_{z}$ - orbital.

4. $4 f$ -orbital have $n=4$ and $l=3$ and the number of radial nodes is $n-l-1=0=4-3-1=0$.