Chemistry Test - 9

Both statements are incorrect.

The electron-pair bond forms through the interaction of unpaired electrons on each of the two atoms.Once paired, the two electrons cannot take part in additional bonds.

Hence option D is the correct answer.

\begin{equation}\underset{(A)}{A g N O_{3}}+\underset{(B)}{NaBr}\rightarrow\underset{(C)}{AgBr}+N a N O_{3}\end{equation} Hence option A is correct.

\(\begin{aligned}(I E)_{H e^{+}}=\left(-E_{n}\right) &=\frac{2.18 \times 10^{-18} Z^{2}}{n^{2}} \text { Jatom }^{-1} \\ &=\frac{2.18 \times 10^{-18} \times(2)^{2}}{1} \text { Jatom }^{-1} \text {for } H e^{+}(n=1, Z=2) \\ &=2.18 \times 10^{-18} \times 4 \times 6.02 \times 10^{23} \mathrm{Jmol}^{-1} \\ &=5.2494 \times 10^{5} \mathrm{Jmol}^{-1} \\ &=5.2494 \times 10^{3} \mathrm{kJmol}^{-1} \end{aligned}\)

Hence option B is the correct answer.

This is because O⁻ ion will resist the addition of another electron. Anion(O⁻) already has a negative electron cloud on itself, and thus, will resist the further addition of a second electron. Hence, energy has to be supplied to overcome the repulsion. Thus, the second step is an endothermic reaction.

Hence option C is the correct answer.

Ba<Ca<Se<S<Ar is the correct order of increasing first ionization enthalpy. Ionization enthalpy increases along the period but decreases down the group. The IE of an element increases as one moves across a period in the periodic table because the electrons are held tighter by the higher effective nuclear charge. The ionization energy of the elements decreases as one moves down the group because the electrons are held in lower-energy orbitals, away from the nucleus and therefore, are less tightly bound. Ar has higher IE because, it is a noble gas and Ba has lowest IE as it is in 6 period and more metallic.

Hence option A is the correct answer.

The highest ionization is possessed by the element which will not want to lose its electron. Consider the options:
A) \([N e] 3 s^{2} 3 p^{1}-\) Can lose the \(p\) electron with relative ease.
B) \([N e] 3 s^{2} 3 p^{3}-\) since it is half filled, it is difficult to remove an electron. Hene ionization energy will be high in this case.
C) \([N e] 3 s^{2} 3 p^{2}-\) Can lose \(p\) electron with relative ease.
D) \([A r] 3 d^{10}, 4 s^{2} 4 p^{3}-\) Half filled. Difficult to remove electron. Hence, we are left with two options, either \(B\) or \(D .\) But in option \(D\) the half filled configuration is in \(4 p\) whereas it is in \(3 p\) in option \(B\). It is relatively easier to remove an electron from \(4 p\) than \(3 p .\)

Effective nuclear charge \(Z_{e f f}=Z-S\)
where \(Z=\) atomic number and \(S=\) screening constant
\(=0.35\) per electron for electron in nth orbit.
\(=0.85\) per electron for electrons in \((n-1)\) th orbit
\(=1.00\) per electron for electrons in \(( n -2) th ,( n -3)\) th, \(( n -4)\) th orbit
\(=0.30\) per electron in \(1 s\) -orbital (when alone)
\(B e=1 s^{2} 2 s^{2}\) one valence-electron in \(2 s\) is screened by one electron in \(2 s-\) orbital (nth orbit) and two electrons in 1 s orbital ( \((n-1)\) th orbit)
\(\therefore S=0.35+2 \times 0.85=2.05\)
\(\therefore Z_{e f f}=4-2.05=1.95\) (given)
\(B e^{+}: 1 s^{2} 2 s^{1}, 2 s\) -electron is screened by two electrons in 1 s-orbital \((( n -1)\) th orbital \()\) \(S=2 \times 0.85=1.70\)
\(\therefore Z_{e f f}=4-1.70=2.30\)
\(B e^{2+}: 1 s^{2}\) one electron in 1 s-orbital (alone exists) is screened by another electron in same orbit. Thus, \(S=0.30\)
\(\therefore Z_{e f f}=4-0.30=3.70\)
\(B e^{3+}: 1 s^{1}\) no-screening (single electron) Thus, \(Z_{e f f}=4\)
Thus, option \(A\) is correct.