Niobium (Nb) and ruthenium (Ru) have “x” and “y” number of electrons in their respective 4 d orbitals. The value of x+y is .
Answer
11
Solution
Nb (Z=41): [Kr] 4d45s1 ⇒ x=4
Ru (Z=44): [Kr] 4d75s1 ⇒ y=7
∴ x+y=11
Answer: 11
Niobium (Nb) and ruthenium (Ru) have “x” and “y” number of electrons in their respective 4 d orbitals. The value of x+y is .
11
Nb (Z=41): [Kr] 4d45s1 ⇒ x=4
Ru (Z=44): [Kr] 4d75s1 ⇒ y=7
∴ x+y=11
Answer: 11
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