Classification of Elements and Periodicity in Properties MCQ Questions & Answers in Inorganic Chemistry | Chemistry
Learn Classification of Elements and Periodicity in Properties MCQ questions & answers in Inorganic Chemistry are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
101.
Which of the following is the correct order of size of the given species ?
A
$$I > {I^ - } > {I^ + }$$
B
$${I^ + } > {I^ - } > I$$
C
$$I > {I^ + } > {I^ - }$$
D
$${I^ - } > I > {I^ + }$$
Answer :
$${I^ - } > I > {I^ + }$$
Cation is smaller and anlon is larger in size than its parent atom.
102.
Which of the following can most easily form unipositive gaseous ion?
A
$$1{s^2}2{s^2}2{p^6}$$
B
$$1{s^2}2{s^2}2{p^6}3{s^2}3{p^1}$$
C
$$1{s^2}2{s^2}2{p^6}3{s^2}$$
D
$$1{s^2}2{s^2}2{p^6}3{s^2}3{p^3}$$
Answer :
$$1{s^2}2{s^2}2{p^6}3{s^2}3{p^1}$$
No explanation is given for this question. Let's discuss the answer together.
103.
The electronegativity follows the order
A
$$F > O > Cl > Br$$
B
$$F > Cl > Br > O$$
C
$$O > F > Cl > Br$$
D
$$Cl > F > O > Br$$
Answer :
$$F > O > Cl > Br$$
$$F$$ 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
A
polarity of the molecules
B
position in the emf series
C
coordination numbers
D
dipole moments
Answer :
polarity of the molecules
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
A
extra stability of half filled orbitals
B
more number of energy levels
C
less number of valence electrons
D
smaller size
Answer :
extra stability of half filled orbitals
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
A
$$52\,J,\,49.5\,J$$
B
$$520J,495\,J$$
C
$$49.5\,J,52\,J$$
D
$$495\,J,520\,J$$
Answer :
$$520J,495\,J$$
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?
A
Principal quantum number $$(n)$$
B
Nuclear charge $$(Z)$$
C
Nuclear mass
D
Number of core electrons
Answer :
Nuclear mass
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.