$$F$$ and $$O$$ belong to 2nd period whereas $$Cl$$ and $$Br$$ belong to 3rd and 4th periods respectively. Hence the sequence of the $$E.N.$$ is $$F > O > Cl > Br$$
104.
Pauling’s electronegativity values for elements are useful in predicting
Pauling’s electronegativity values are useful in determination of polarity of the bond in molecules. If electronegativity difference is zero, then the molecule is non-polar otherwise it is polar.
$${X_A} - {X_B} = 0.028\sqrt \vartriangle E$$
$${X_A}$$ and $${X_B}$$ are electronegativities of the atoms $$A$$ and $$B$$ respectively. While,
$$\vartriangle E = \,{\text{actual}}\,{\text{bond}}\,{\text{energy}}$$ $$ - \sqrt {{E_{A - A}} \times {E_{B - B}}} $$
105.
Ionisation enthalpy of nitrogen is more than oxygen because of
Ionisation enthalpy of nitrogen is more than oxygen due to extra stability of half filled orbitals.
106.
The ionization enthalpies of $$Li$$ and $$Na$$ are $$520\,kJ\,mo{l^{ - 1}}$$ and $$495\,kJ\,mo{l^{ - 1}}$$ respectively. The energy required to convert all the atoms present in $$7\,mg$$ of $$Li$$ vapours and $$23\,mg$$ of sodium vapours of their respective gaseous cations respectively are
No. of moles of $$Li = \frac{7}{{1000 \times 7}} = {10^{ - 3}}$$
No. of moles of $$Na = \frac{{23}}{{1000 \times 23}} = {10^{ - 3}}$$
∴ The amount of energies required for $${10^{ - 3}}$$ mole each of $$Li$$ and $$Na$$ are $$520\,kJ \times {10^{ - 3}}$$ and $$495\,kJ \times {10^{ - 3}}$$ or $$520\,J$$ and $$495\,J$$ respectively.
107.
The first $$\left( {I{E_1}} \right)$$ and second $$\left( {I{E_2}} \right)$$ ionisation enthalpies $$\left( {kJ/mole} \right)$$ of element, $$A$$ are given below.
$$A$$
$$I{E_1}$$
750
$$I{E_2}$$
1500
In an experiment $$\frac{1}{{12}}\,mole$$ of $$A$$ atom in vapour phase absorb $$100\,kJ$$ of energy. So that it forms a mixture of $${A^ + }$$ and $${A^{2 + }}$$ ions. Energy utilized to form $${A^{2 + }}$$ from $${A^ + }$$ in experiment is
In isoelectronic species the number of electrons are same but nuclear charge is different. As the nuclear charge increase, the attraction force on last electron increases, so the size decreases or in other words
\[\text{Ionic}\,\,\text{size}\propto \frac{1}{\text{Charge}\,\,\text{on}\,\,\text{cation}}\] and hence, order is
110.
Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?
Nuclear mass does not affect the valence shell because nucleus consists of protons and neutrons.
Protons i.e., nuclear charge affect the valence shell but neutrons do not.