Question

The orbital having n=2,1=1 and m=0 is designated as (a) 2pz (b) 2px (c) 2py
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Answer 1

Answer:

The orbital having $n=2, l=1 ,m=0$ is designed as $2P_z$

Explanation:

Given the quantum numbers of an orbital as $n=2, l=1 ,m=0$

$n$ is the principal quantum number. Here, $n=2$  so it is 2nd orbital

l is the azimuthal quantum number,  $l=0$ for s orbital, $l=1$ for p orbital, $l=2$ for d orbital. Here,  $l=1$  so it is p orbital

m is the magnetic quantum number, P orbital degenerate as $P_x, P_y,P_z$ for $m=0$ it is $P_z$