For an orbital in `B^(+4)` radial function is :
`R(r ) = (1)/(9sqrt(6))((z)/(a_(0)))^((3)/(4))(4-sigma)sigma e^(-sigma//2`
where `sigma = (Zr)/(a_(0))` and `a_(0)=0.529Å,Z` = atomic number, `r=` radial distance from nucleus.
The radial node of orbital is at distance from nucleous.

Options

A. `0.529Å`
B. `2.12Å`
C. `1.06Å`
D. `0.423Å`

Explanation

To find the radial node of the orbital in \( B^{+4} \), we start with the given radial function:

\[ R(r) = \frac{1}{9\sqrt{6}} \left( \frac{Z}{a_0} \right)^{\frac{3}{4}} (4 - \sigma) \sigma e^{-\frac{\sigma}{2} \]

where \( \sigma = \frac{Zr}{a_0} \) and \( a_0 = 0.529 \, \text{Å} \).

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