Question

Correct order of radius of the 1st orbit of H, $H{{e}^{+}}$, $L{{i}^{2+}}$ and $B{{e}^{3+}}$ is:
(A)- $H>H{{e}^{+}}>L{{i}^{2+}}>B{{e}^{3+}}$
(B)- $B{{e}^{3+}}>L{{i}^{2+}}>H{{e}^{+}}>H$
(C)- \[H{{e}^{+}}>B{{e}^{3+}}>L{{i}^{2+}}>H\]
(D)- $H{{e}^{+}}>H>L{{i}^{2+}}>B{{e}^{3+}}$

Answer 1

Hint: From the Bohr’s model of hydrogen atom, the expression for radius of the ${{n}^{th}}$orbit of the hydrogen atom is given as
     \[r={{a}_{o}}{{n}^{2}}\]
${{a}_{o}}$ is the radius of the first orbit of the hydrogen atom and is equal to 52.9 pm.
For H-like particles, radius of the ${{n}^{th}}$orbit is given by the expression
     \[r=\dfrac{{{a}_{o}}{{n}^{2}}}{Z}\]

Complete answer:
$H{{e}^{+}}$, $L{{i}^{2+}}$ and $B{{e}^{3+}}$contain one electron like hydrogen, so they are called H-like particles.
Radius of the first orbit of the hydrogen atom is called Bohr's radius, i.e. ${{a}_{o}}=52.9pm$.
The expression for the radius of ${{n}^{th}}$ orbit for H-like particles is given as
     \[r=\dfrac{{{a}_{o}}{{n}^{2}}}{Z}\]
Where, Z is the atomic number of H-like particles.
We have to find the order of the radius of the 1st orbit, i.e. n = 1.
Now, the radius of the first orbit for $H{{e}^{+}}$, $L{{i}^{2+}}$ and $B{{e}^{3+}}$ becomes
     \[r=\dfrac{{{a}_{o}}}{Z}\]
Since, the value of ${{a}_{o}}$ is constant, i.e. 52.9 pm, therefore, we can say that radius r is inversely proportional to the atomic number Z of the H-like particle, i.e.
     \[r\propto \dfrac{1}{Z}\]
Atomic number, Z for $H{{e}^{+}}$, $L{{i}^{3+}}$ and $B{{e}^{3+}}$ is 2, 3 and 4, respectively.
Since, atomic number Z increases from H to $B{{e}^{3+}}$, radius r being inversely proportional decreases from $B{{e}^{3+}}$ to H. This implies that the radius of the first orbit is largest for H and smallest for $B{{e}^{3+}}$.
Therefore, the correct order of radius of the 1st orbit of H, $H{{e}^{+}}$, $L{{i}^{2+}}$ and $B{{e}^{3+}}$ is:
     $H>H{{e}^{+}}>L{{i}^{2+}}>B{{e}^{3+}}$

Hence, the correct option is (A).

Note:
As the atomic number increases from H to $B{{e}^{3+}}$, the effective nuclear charge on the electron increases. This means that the electron is now strongly attracted by the nucleus which results in the contraction of the size and hence, the radius decreases from H to $B{{e}^{3+}}$.