MCQ Questions for Class 11 Medical Chemistry Classification Of Elements And Periodicity In Properties Quiz 2

$CuX_2$ , X belongs to which group?

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group IA
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group IIA
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group IIIA
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group VIA
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group VIIA

Explanation

$CuX_2$ , as the O.S. of

$Cu$ is

$+2$ , the O.S of

$X$ is

$-1$ , hence the

$X$ belongs to group VIIA.

Which of the following has the lowest ionization energy?

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$Li$
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$Na$
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$K$
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$Rb$
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$Cs$

Which element reacts most explosively in water?

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$K$
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$Ca$
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$Br$
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$Kr$
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$Zn$

Explanation

Potassium (K) reacts most explosively in water.
Except lithium, all alkali metals reacts explosively with water.

$\displaystyle 2K +2H_2O \rightarrow 2K^+ + 2OH^- + H_2\uparrow$

Arrange the elements in order of their reactivity?
B, Al, Ga, In

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B < Al < Ga < In
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In < Ga < Al < B
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Ga < In < Al < B
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None of the above

Which statement describes positive ions?

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Positive ions have more electrons than neutrons
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Positive ions have more protons than neutrons
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Positive ions have more electrons than protons
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Positive ions have more protons than electrons

The electronic configuration of the two outermost shells of an atom is

$3s^23p^63d^54s^2$ . What is this atom?

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Manganese
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Phosphorus
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Strontium
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Vanadium

Explanation

$3s^23p^63d^54s^2$

$\text{ is the outermost electronic configuration of Manganese.}$

Electronic configuration of

$H^{ - }$ is:

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$1 s ^ { 0 }$
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$1 s ^ { 1 }$
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$1 s ^ { 2 }$
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$1 s ^ { 1 }$ $2 s ^ { 1 }$

Which of the following gas does not have an octet or eight electrons in the outer shell

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$Ne$
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$Ar$
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$Rn$
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$He$

Explanation

$He$ has

$2$ electrons in outermost shell.

An element 'X' with atomic number 114 has recently been discovered. Its IUPAC name is:

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Eka-Iead
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Ununfortium
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Ununquadium
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Bohrium

Explanation

$Un - un -quad -ium$

$1$

$4$

Ununquadium is the IUPAC name of the elements with atomic number 114.

Configuration of the final elements of

$3d, 4d, 5d$ series respectively are:

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$3d^{5}4s^{2}, 4d^{10}5s^{2}, 5d^{10}6s^{2}$
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$3d^{10}4s^{2}, 4d^{10}5s^{2},5d^{10}6s^{2}$
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$3d^{10}4s^{2}, 4d^{10}5s^{2}s5d^{10}6s^{1}$
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$3d^{9}4s^{2},4d^{10}5s^{2},5d^{10}6s^{1}$

General outermost electronic configuration of the elements situated above and below

$X$ (Atomic number

$=50$ ) would be:

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$ns^{2}np^{2}$
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$Ns^{2}np^{3}$
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$ns^{2}np^{4}$
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$ns^{2}np^{5}$

The first ionisation potential is maximum for :

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lithium
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uranium
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iron
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hydrogen

The positions of four elements A, B, C and D in the modern periodic table are shown below. Which element is most likely to form an acidic oxide?

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A
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B
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C
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D

Explanation

$\textbf{Correct option:}$ (C)

$\bullet$ The metals are a group of elements located on left side of the periodic table and non- metals are a group of elements located on the right side of the periodic table (except for hydrogen, which is on the top left).

$\bullet$ Noble gas are present in the last column of the peridic table.

$\textbf{Explanation for correct option:}$

$\textbf{(C)}$

$\bullet$ Element C is a non- metal.

$\bullet$ Non- metal oxides react with water to form acidic solutions which liberates Hydrogen ions in the solution.

$\bullet$ Therefore, non-metals form acidic oxides which is element C.

$\textbf{Explanation for incorrect option:}$

$\textbf{(A)}$

$\bullet$ Element A is a metal.

$\bullet$ Metal forms basic oxides.

$\textbf{(B)}$

$\bullet$ Element B is a metal.

$\bullet$ Metal forms basic oxides.

$\textbf{(D)}$

$\bullet$ Element D is a noble gas.

$\bullet$ Noble gas will not form any kind of oxide.

$\textbf{Conclusion:}$

Hence, the correct option is (C).

The properties of _________ were predicted by Mendeleev before their isolation.

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Co and Ni
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I and Te
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Sc, Ga and Ge
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Cl, Ar and K

Explanation

Mendeleev has the foresight to leave some gaps in the periodic table for 3 elements and these elements are discovered later and included in the table.

Those three elements are:
1) Eka-boron is presently known as scandium (

$\text{Sc}$ ).
2) Eka-silicon is presently known as germanium (

$\text{Ge}$ ).
3) Eka-aluminium is presently known as gallium(

$\text{Ga}$ ).

Hence, the correct option is

$\text{C}$

In Mendeleev table, the triad of VIII group is :

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Ru, Rh, Pd
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Cu, Ag, Au
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N, O, F
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Tl, Pb, Bi

Explanation

Group VIII of Mendeleev's table consists of three triads known as transition triads and they are
i) Iron, Cobalt, and Nickel
ii) Ruthenium, Rhodium, and Palladium
iii) Osmium, Iridium, and Platinium

Option

$A$ is correct.

Which of the following electronic configurations is characteristic of alkali metals?

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$(n-1)p^{6},ns^{2},np^{1}$
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$( n- 1 )s^{2}p^{6}d^{10},ns^{1}$
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$( n-1)p^{6},ns^{1}$
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$ns^{2},np^{6},nd^{1}$

Explanation

The outer most shell electronic configuration of alkali metals (IA group elements) is

$ns^1$ . s-block is filled after p-block of the penultimate shell. Some of the electronic configurations of alkali metals are as follows:

$Li= 1s^22s^1$

$Na= 1s^22s^22p^63s^1$

$K= 1s^22s^22p^63s^23p^64s^1$

$Rb= 1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^1$

Mendeleev corrected the atomic weight of :

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$Be$
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$N$
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$O$
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$Cl$

Identify the property which does not reflect the periodicity of the elements.

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Bonding behaviour
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Electronegativity
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Ionization potential
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Neutron-proton ratio

A pair of atomic numbers which belong to s-block are:

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7, 15
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6, 12
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9, 17
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3, 12

The correct order of the second ionisation potential of carbon, nitrogen, oxygen and fluorine is :

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$C>N>O>F$
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$O>N>F>C$
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$O>F>N>C$
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$F>O>N>C$

Explanation

The electronic configurations of ions after first ionization are:

$O^+: 1s^22s^22p^3$

$F^+: 1s^22s^22p^4$

$N^+: 1s^22s^22p^2$

$C^+: 1s^22s^22p^1$
As we go across a period, Zeff increases because valence electrons do not screen the nuclear charge effectively. Hence, the IP's increase across a period. But, oxygen has more second ionisation potential than fluorine. This is because after the first ionisation, oxygen acquires stable half filled configuration. Thus, for the second ionisation, electron has to be removed from

$2p^3$ sub-shell. While in the case of fluorine, it is getting removed from

$2p^4$ sub-shell.

Number of short periods in Mendeleev's periodic table is :

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$2$
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$3$
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$4$
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$1$

The ionization energy of nitrogen is more than that of oxygen because :

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of the extra stability of half-filled p orbitals in nitrogen
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of the smaller size of nitrogen
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the former contains less number of electrons
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the former is less electronegative

Explanation

Ionization energy is the minimum energy required to remove the most loosely held electron from one mole of gaseous atoms to produce one mole of gaseous ions each with a charge of

$+1$ .

An atom is highly stable when its valence orbital is completely filled or half-filled and in nitrogen, the valence p orbital is exactly half filled. So, p orbital in nitrogen is more stable than in oxygen, which has one electron more than the half-filled configuration. So, the ionization energy of nitrogen is more than that of oxygen.

Considering the chemical properties, atomic weight of

$Be$ was corrected based on:

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electronic configuration
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valency
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atomic number
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both B and C

Explanation

The atomic weight of Be was corrected based on:

Atomic weight

$=$ equivalent weight

$\times$ valency

From this, we can understand that the atomic weight of

$Be$ was corrected based on valency.

Ionisation potential of boron is less than that of beryllium. This is because :

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B has $1s^{2} 2s^{2} 2p^{1}$ configuration
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B has small atomic size
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B has higher nuclear charge
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B has more number of shells

Which has least ionization potential?

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$Li$
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$Cs$
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$Cl$
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$I$

Explanation

Ionization potential increases on going from left to right in a period and decreases down the group. Thus, I.P order is

$Cs<Li < I<Cl$ .

The ionization potential of elements in any group decreases from top to bottom. This is due to :

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increase in the size of an atom
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increase in the atomic number
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increase in the screening effect
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both increase in the size of an atom and increase in the screening effect

The I.P. of sodium is 5.14 eV. The I.P. of potassium could be :

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same as that of sodium
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5.68 eV
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4.34 eV
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10.28 eV

Explanation

I.P decreases on going down the group as size increases and the outer electrons experience less effective nuclear charge. Therefore, I.P of

$K<Na$ .

The electron configuration of elements A, B and C are

$[He] 2s^{1},\ [Ne]3s^{1}$ and

$[Ar] 4s^{1}$ respectively. Which one of the following order is correct for the first ionization potentials

$(in\ KJ\ mol^{1})$ of A, B and C?

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A > B > C
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C > B > A
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B > C > A
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C > A > C

Explanation

Ionization potential decrease down the group as the attraction over outer electrons decreases. Hence, correct order is

$A>B>C$ .

The high ionistion potential of magnesium compared with aluminium is due to :

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filled orbitals in magnesium
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high nuclear charge in magnesium
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low radius of magnesium atom
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low effective nuclear charge in magnesium

Explanation

Magnesium has a stable full-filled configuration. It is difficult to remove an electron from such a configuration.

Thus, I.P of

$Mg$ is greater than that of

$Al$ .

The ionisation potential is very low for :

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$Be$
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$Mg$
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$B$
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$Al$

Elements X, Y, and Z have atomic numbers

$19, 37$ , and

$55$ respectively. Which of the following statements are true about them?

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Their ionization potential would increase with increasing atomic number
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Y would have an ionization potential between those of X and Z
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Z would have the highest ionization potential
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Y would have the highest ionization potential

Explanation

Elements X , Y and Z with atomic numbers

$19, 37 , 55$ lie in group

$1$ (alkali metals). Within a group, IE decreases from top to bottom. Therefore, IE of Y could be between those of X and Z.

The

$IP_{1}$ value of potassium is less than the

$IP_{1}$ value of sodium. This is due to :

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large size of potassium atom
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small size of potassium atom
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low density of potassium
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univalent nature of potassium

Explanation

The outer electron in potassium is in the

$4s$ orbital, which is further away from the nucleus than the

$3s$ orbital of the sodium.

The greater distance means that the attraction between the electron is weaker.

The electron in Potassium is also more affected by shielding due to more shells, further weakening this attraction. This means that less energy is needed to remove the outermost electron and therefore the ionisation energy is lower.

Therefore,

$IP_1$ value of

$K$ <

$IP_1$ value of

$Na$ .

Hence, option A is correct.

The set representing the correct order of first ionisation potential is :

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$K>Na>Li$
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$Be>Mg>Ca$
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$B>C>N$
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$Ge>Si>C$

Explanation

Size increases as we move down the group. Therefore, the outer electrons experience less attraction due to increase in distance, i.e., the effective nuclear charge experienced is less. Less the effective nuclear charge less the I.P. Therefore, I.P of

$Be>Mg>Ca$ .

The incorrect statement among the following is :

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the first ionisation potential of $Al$ is less than the first ionisation potential of $Mg$
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the second ionisation potential of $Mg$ is greater than the second ionisation potential of $Na$
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the first ionisation potential of $Na$ is less than the first ionisation potential of $Mg$
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the third ionisation of potential of $Mg$ is greatest

Beryllium shows diagonal relationship with aluminium. Which of the following similarity is incorrect?

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$Be_{2}C$ like $Al_{4}C_{3}$ yields methane on hydrolysis
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$Be$ , like $Al$ is rendered passive by $HNO_{3}$
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Be $(OH)_{2}$ like $Al(OH)_{3}$ is basic
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$Be$ forms beryllates and $Al$ forms aluminate

Explanation

Diagonal relationship of Be with Al is because of small size. Be is different from other earth alkaline earth metals. But it resembles in many of its properties with Al. The $Be(OH)_{2}$ and $Al(OH)_{3}$ are amphoteric in nature. The two structures involve the only movement of electrons and not of atoms or groups, hence these are resonating structures.

Hence, the correct answer is option $\text{C}$ .

The decreasing order of the second ionization potential of

$K, \ Ca$ and

$Ba$ is :

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$K > Ca > Ba$
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$Ca > Ba > K$
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$Ba > K > Ca$
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$K > Ba > Ca$

Diagonal relationship is quite pronounced in the elements of:

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$2^{nd}$ and $3^{rd}$ periods
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$1^{st}$ and $2^{nd}$ periods
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II and III groups
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$3^{rd}$ and $4^{th}$ periods

The decreasing order of electron affinity of halogens is :

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$F > Cl > Br > I$
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$F < Cl < Br < I$
0%
$F < Cl > Br < I$
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$Cl > F > Br > I$

Explanation

Electron affinity increases in a period and decreases in a group. But, fluorine atom, due to its small size cannot release high amount of energy. Hence, among halogens, chlorine atom has high value of electron affinity and the order is

$Cl>F>Br>I$ .

Boron is diagonally related to:

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boron
0%
aluminium
0%
magnesium
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silicon

Among the following four elements, the ionisation potential of which one is the highest?

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Oxygen
0%
Argon
0%
Barium
0%
Caesium

Boron and silicon resemble chemically. This is due to the equal value of their:

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electron affinity
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atomic volume
0%
ions polarizing power
0%
nuclear charge

Diagonal relationship is shown by

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$B - S$
0%
$Li - Mg$
0%
$Mg - Ca$
0%
$S - Se$

Explanation

Hint:

Due to the identical size of ions, a diagonal relationship is formed. As we move towards right in the periodic table, then the size of an atom gradually decreases, and as we move downward across a group, then the size of the atom gradually increases.

Explanation:

As we move across the period, the size of atoms decreases and as we move down the group, the size of the atoms gradually increases. The diagonal relationship is due to the similarity in size of the atoms.

Diagonal relationship in the periodic table.

For example, Boron and silicon are both semiconductors that generate halides that are hydrolysed in water and have acidic oxides.

Therefore, Diagonal relationship is shown by $Li-Mg$ .

Final Answer is Option (B).

Which of the following will have high ionization potential?

0%
$Al^{3+}$
0%
$F^{-}$
0%
$Na^{+}$
0%
$O^{2-}$

Explanation

$Al^{3+} , Na^{+}, O^{2-}$ and

$F^{-}$ are iso-electronic. I.P is directly proportional to effective nuclear charge. Effective nuclear charge is directly proportional to positive charge. Thus, I.P order is :

$Al^{3+} > Na^{+}> F^{-} >O^{2-}$ .

If the ionization potential of calcium is five units then that of magnesium may be :

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6 Units
0%
5 Units
0%
4 Units
0%
2 Units

$Al$ has lower ionization potential than that of

$Mg$ because :

0%
$Al$ atom is bigger than $Mg$ atom
0%
$Mg$ atom is bigger than $Al$ atom
0%
all electrons in $Mg$ are paired, but those of $Al$ are not
0%
all belongs to a higher group

Explanation

Ionization potential decreases on going from left to right but ionization potential of

$Al < Mg$ . Aluminium, after losing one electron, attains a stable full filled configuration and magnesium has a stable full filled configuration. Therefore, it is easier to remove an electron from aluminium and difficult to remove from magnesium. Thus, ionization potential of

$Mg > Al$ .

The first I.P. values in electron volts of nitrogen and oxygen atoms are respectively given by :

0%
$14.6, \ 13.6$
0%
$13.6,\ 14.6$
0%
$13.6,\ 13.6$
0%
$14.6,\ 14.6$

As you go down in a group, the alkali metals become:

0%
brighter
0%
hotter
0%
more reactive
0%
less reactive

The following are some statements about diagonal relationship.
i) It is due to nearly same size, electronegativity and same polarising power
ii)

$Li$ has similar properties with

$H_{2}$
iii)

$BeO$ ,

$Al_{2}O$ are amphoteric in nature
iv) All

$d$ -block elements are stable

0%
Only (i) correct
0%
Only (ii) correct
0%
Only (iii) correct
0%
Only (iv) is correct

The ionization potential

$(I_{1})$ of nitrogen (

$Z=7$ ) is more than oxygen (

$Z=8$ ). This is explained by :

0%
Hunds rule
0%
Excitation rule
0%
Pauli principle
0%
Aufbau principle

The correct order of ionization energies is :

0%
$Zn < Cd < Hg$
0%
$Hg < Cd < Zn$
0%
$Ar > Ne > He$
0%
$Cs < Rb < Na$

Explanation

Hint: Ionization energy decreases down the group with an increase in size.

Correct Option: $D$

Explanation of correct option:

Ionisation energy is the minimum amount of energy required to remove a valence electron from an isolated gaseous atom.

As we move across the period, the atomic number of the atom increases and simultaneously the number of electrons increases in the same valence shell.

On moving down in a group, the valence shell becomes far away from the nucleus and thus nucleus attraction towards test electron decreases which results in a decrease in ionization energy.

Therefore, the correct order of ionization energies of given elements will be:

$Cs<Rb<Na$

Explanation of incorrect option:

Only

$s$ -group elements can easily lose their outer shell electrons at low ionization energy, hence all other options are incorrect.

CBSE Questions for Class 11 Medical Chemistry Classification Of Elements And Periodicity In Properties Quiz 2 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Classification Of Elements And Periodicity In Properties Quiz 2 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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