number of angular nodes possible in the outermost orbital of sodium
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Electronic configuration of:
Na= 1s2 2s2 2p6 3s1
Number of angular nodes = n-l-1
n= 3 l= 0 so angular nodes= 3-1= 2
Na= 1s2 2s2 2p6 3s1
Number of angular nodes = n-l-1
n= 3 l= 0 so angular nodes= 3-1= 2
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