Medium
The electronic configurations of four elements are given in brackets: $L\,(1s^2, 2s^2 2p^1)$; $M\,(1s^2, 2s^2 2p^5)$; $Q\,(1s^2, 2s^2 2p^6, 3s^1)$; $R\,(1s^2, 2s^2 2p^2)$. Which element would most readily form a diatomic molecule?
- A$Q$
- B$M$
- C$R$
- D$L$
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Similar Questions
Which one is the electronic configuration of $Fe^{2+}$?
The number of radial and angular nodes in $4d$ orbital are,respectively $........$ .
The electronic configuration $1s^2 2s^2 2p^6 3s^2 3p^6 3d^6$ corresponds to:
Easy
View SolutionMatch the column:
Column-$I$ Column-$II$ $A$. The orbital which has two angular nodes $P$. $4d_{x^2-y^2}$ $B$. The $d$ orbital with zero nodal plane $Q$. $3d_{z^2}$ $C$. The orbital with two radial nodes $R$. $4f$ $D$. The orbital with three angular nodes $S$. $3s$
Correct match will be $:-$
| Column-$I$ | Column-$II$ |
|---|---|
| $A$. The orbital which has two angular nodes | $P$. $4d_{x^2-y^2}$ |
| $B$. The $d$ orbital with zero nodal plane | $Q$. $3d_{z^2}$ |
| $C$. The orbital with two radial nodes | $R$. $4f$ |
| $D$. The orbital with three angular nodes | $S$. $3s$ |
Correct match will be $:-$
Difficult
View SolutionMatch the quantum numbers with the information provided:
Quantum Number Information $(1)$ Principal quantum number $(A)$ Spatial orientation of the orbital $(2)$ Azimuthal quantum number $(B)$ Energy and size of the orbital $(3)$ Magnetic quantum number $(C)$ Electron spin $(4)$ Spin quantum number $(D)$ Shape of the orbital
| Quantum Number | Information |
| $(1)$ Principal quantum number | $(A)$ Spatial orientation of the orbital |
| $(2)$ Azimuthal quantum number | $(B)$ Energy and size of the orbital |
| $(3)$ Magnetic quantum number | $(C)$ Electron spin |
| $(4)$ Spin quantum number | $(D)$ Shape of the orbital |
Medium
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