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Assertion: Mn atom loses ns electrons fi...

Assertion: Mn atom loses ns electrons first during ionisation as compared to `(n-1)` d electrons
Reason: The effective nuclear charge experienced by `(n-1)`d electrons is greater than that by ns electrons.

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R is the correct explanation of A.
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The reducing effect of the nuclear charge by the inner electrons for on outer electron is termed aas shielding (or screening). As a result of shielding, the outer electrons in an atom always experience less nnuclear charge than the actual nuclear charge Z. The effective nuclear charge (Z^(**)) as experienced by an electron is then obtained by subtracting the total shielding contributions from alll the other electrons (i.e., except the one under consideration) from the actual nuclear charge. Z^(**)=Z-sigma Where sigma =sum of the shielding contributions. The rules for estimating contributions to sigma are as follows (Slater's rule) Contribution to shielding by each electron is : |{:("Electron","All Higher","Same","Group","Group"le),("Grpoup","Group","Group",n-1,n-2),(1s," "0,0.30,-,-),((ns,sp)," "0,0.35,0.85,1.00),((nd)or(nf)," "0,0.35,1.00,1.00):}| According to Slater's treatment, the energy of an electron in nth shell of an atom having atomic number Z is given by the empirical equation E=-13.6((Z^(**))/(n))^(2)eV Z^(**) = effective nuclear charge The magnitude of first ionisation energy for Na (according to formula given) is equal to:

The reducing effect of the nuclear charge by the inner electrons for on outer electron is termed aas shielding (or screening). As a result of shielding, the outer electrons in an atom always experience less nnuclear charge than the actual nuclear charge Z. The effective nuclear charge (Z^(**)) as experienced by an electron is then obtained by subtracting the total shielding contributions from alll the other electrons (i.e., except the one under consideration) from the actual nuclear charge. Z^(**)=Z-sigma Where sigma =sum of the shielding contributions. The rules for estimating contributions to sigma are as follows (Slater's rule) Contribution to shielding by each electron is : |{:("Electron","All Higher","Same","Group","Group"le),("Grpoup","Group","Group",n-1,n-2),(1s," "0,0.30,-,-),((ns,sp)," "0,0.35,0.85,1.00),((nd)or(nf)," "0,0.35,1.00,1.00):}| According to Slater's treatment, the energy of an electron in nth shell of an atom having atomic number Z is given by the empirical equation E=-13.6((Z^(**))/(n))^(2)eV Z^(**) = effective nuclear charge Among the following, which electron of Fe atom experience minimum attraction from nucleus? (Atomic number of Fe = 26)

The reducing effect of the nuclear charge by the inner electrons for on outer electron is termed aas shielding (or screening). As a result of shielding, the outer electrons in an atom always experience less nnuclear charge than the actual nuclear charge Z. The effective nuclear charge (Z^(**)) as experienced by an electron is then obtained by subtracting the total shielding contributions from alll the other electrons (i.e., except the one under consideration) from the actual nuclear charge. Z^(**)=Z-sigma Where sigma =sum of the shielding contributions. The rules for estimating contributions to sigma are as follows (Slater's rule) Contribution to shielding by each electron is : |{:("Electron","All Higher","Same","Group","Group"le),("Grpoup","Group","Group",n-1,n-2),(1s," "0,0.30,-,-),((ns,sp)," "0,0.35,0.85,1.00),((nd)or(nf)," "0,0.35,1.00,1.00):}| According to Slater's treatment, the energy of an electron in nth shell of an atom having atomic number Z is given by the empirical equation E=-13.6((Z^(**))/(n))^(2)eV Z^(**) = effective nuclear charge Z^(**) for a 1st electron in Fe atom is:

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