The order of screening effect among $s, p, d$ and $f$-orbitals of a given shell of an atom in its outer…
The order of screening effect among $s, p, d$ and $f$-orbitals of a given shell of an atom in its outer shell electrons is
$s>p>d>f$
$f>d>p>s$
$p < d < s>f$
$d>f>p>s$
Solution
$s$ has the greatest effect to reduce the force of attraction between the outermost electron and nucleus due to its effective charge density followed by $p>d>f$.
So, according to the screening effect, $s$ orbital being close to the nucleus lacks screening from any other orbital as it is the first and the closest orbital. $p$ orbital electrons are loosely bound with greater size than $s$ block elements as they undergo screening from $s$ orbital electrons.
Similarly, $d$-orbital electrons undergo screening from $s$ and $p$-block orbitals and finally $f$ orbital also undergoes screening from all the other orbitals. Hence, the $f$-block elements are generally big in size with less nuclear charge acting on their outermost electrons.
Thus, the order of screening effect of electrons of $s, p, d$ and $f$ orbitals of a given shell of an atom on its outer shell electrons is $s>p>d>f$.
Hence, the correct option is (1).