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Be's 4th electron will have four quantum numbers:
n l m s
(1) 1 0 0 +1/2
(2) 1 1 +1 +1/2
(3) 2 0 0 - 1/2
(4) 2 1 0 +1/2
Ans: 2 0 0 - 1/2
NaOH + K2Cr2O7 ---> HCo + NaHCO3 ---> HCo + NaOH ---> CuSo4 + NaOH ---> CH3COOH + NaHCO3 ---> PbNO3 + KI --->
Q. Complexation Behaviour of the f-block? Ions of actinide and lanthanide elements have a strong tendency to form complexes with a variety of oxygen and nitrogen donor ligands.
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