MCQ Questions for Class 11 Medical Chemistry Classification Of Elements And Periodicity In Properties Quiz 12

Which of the following statements is not correct for the element having electronic structure

$2,8,8,2$ ?

0%
Its valency is $2$
0%
It is in fourth period
0%
It is in group $II$
0%
Its atom forms negative ion

Explanation

The element having electronic configuration as

$2,8,8,2$ has valency of

$2$ . It has tendency of losing two electrons and thus, it forms a positive ion.

Which of the following elements will have highest ionization energy ?

0%
$1s^2 2s^2 2p^6 3s^1$
0%
$1s^2 2s^2 2p^6 3s^2 3p^3$
0%
$1s^2 2s^2 2p^6 3s^2 3p^4$
0%
$1s^2 2s^2 2p^6 3s^1$

Identify the incorrect match.

	Name		IUPAC Official Name
a)	Unnilunium	i)	Mendelevium
b)	Unniltrium	ii)	Lawrencium
c)	Unnilhexium	iii)	Seaborgium
d)	Unununnium	iv)	Darmstadtium

0%
(b), (ii)
0%
(c), (iii)
0%
(d), (iv)
0%
(a), (i)

Explanation

The IUPAC name for

$Unununnium$ (

$Z= 111$ ) is

$Roentgenium$ not

$Darmstadtium$

All other matches are correct.

Option C is correct

State True or False:
The first ionization energy of N is lower than that of oxygen.

0%
True
0%
False

Explanation

False, first ionization energy of

$N$ is higher than that of

$O$ .

In group 1 of alkali metals , the ionization potential decreases down the group. therefore , lithium is powerful reducing agents.

0%
True
0%
False

Which of the following has maximum ionization potential ?

0%
Be
0%
K
0%
Na
0%
Mg

Explanation

Because of smallest size,

$Be$ has the highest ionization potential.

The first ionization energy

$(kJ\,mol^{-1})$ for H, Li, F, Na has one of the following values :

$1681, 520, 1312, 495$ .Which of these values corresponds to that of hydrogen?

0%
$1681$
0%
$1312$
0%
$520$
0%
$495$

In which of the following cases, the former change would require lesser energy than the latter?

0%
$Be^{+}\left ( g \right )\rightarrow Be^{2+}\left ( g \right )+e^{-}; Be \left ( g \right )\rightarrow Be^{+}\left ( g \right )+e^{-}$
0%
$Na\left ( g \right )\rightarrow Na^{+}\left ( g \right )+e^{-}; Cl^{-}\left ( g \right )\rightarrow Cl\left ( g \right )+e^{-}$
0%
$Ag\left ( g \right )\rightarrow Ag^{+}\left ( g \right )+e^{-};Au\left ( g \right )\rightarrow Au^{+}\left ( g \right )+e^{-}$
0%
$O^{-}\left ( g \right )\rightarrow O\left ( g \right )+e^{-};S^{-}\left ( g \right )\rightarrow S\left ( g \right )+e^{-}$

Explanation

(A) For all elements,

$IE_{1}< IE_{2}< IE_{3}$

(B) The minimum value of IE for an element in periodic table

$(Na)$ is also greater than the maximum value of

$\Delta H_{eg}$ for an element in periodic table

$(Cl)$ (w.r.t. magnitude only).

$IE$ for

$Au> IE$ for

$Ag$ .

(D)

$\Delta H_{eg}$ for

$S> \Delta H_{eg}$ for oxygen because of small size of oxygen (w.r.t. magnitude only).

Beryllium and aluminium exhibit many properties which are similar. But, the two elements differ in:

0%
maximum covalency in compounds
0%
forming polymeric hydrides
0%
forming covalent halides
0%
exhibiting amphoteric nature in their oxides

Explanation

Beryllium and aluminum exhibit many properties that are similar. But, the two elements differ in exhibiting maximum covalency in compounds.
For example

$Be$ forms

$BeCl_2$ and

$Al$ forms

$AlCl_3$ .

$Be$ and

$Al$ shows some similarities due to the diagonal relationship. Some of these include the following.

They form polymeric hydrides.
Their halides are covalent. Both

$Be$ and

$Al$ form covalent

compounds.

Their oxides are amphoteric in nature.

$BeO+2HCl \rightarrow BeCl_2 + H_2O$

$BeO + 2NaOH \rightarrow Na_2BeO_2+H_2O$

$Al_2O_3+6HCl \rightarrow 2AlCl_3 +3H_2O$

$Al_2O_3 +2NaOH \rightarrow 2NaAlO_2 + H_2O$

Among

$Al_2O_3, SiO_2, P_2O_3$ and

$SO_2$ the correct order of acid strength is:

0%
$Al_2O_3 < SiO_2 < SO_2 < P_2O_3$
0%
$SiO_2 < SO_2 < Al_2O_3 < P_2O_3$
0%
$SO_2 < P_2O_3 < SiO_2 < Al_2O_3$
0%
$Al_2O_3 < SiO_2 < P_2O_3 < SO_2$

Explanation

As electronegativity difference between element and oxygen decreases the acidic character of oxides increases. The electronegativity also increases with increasing oxidiation state. In general, as non-metallic character increases across the period, the acidic character of their oxides increases.

Thus the correct order of acidic strength can be given as:

$Al_2O_3<SiO_2<P_2O_3<SO_2$

IP values and EA values of elements A, B, C and D are given in a tabular form. Identify the strongest oxidizing agent and reducing agent among them.

	IP Value	EA Value
A	450	140
B	225	700
C	300	400
D	700	1035

0%
B, D
0%
B, A
0%
D, A
0%
D, B

Explanation

Electron affinity is the amount of energy released when an electron is added to a neutral atom or molecule in the gaseous state to form a negative ion. More the Electron affinity stronger is the oxidising agent.

Ionization potential is defined as the amount of energy required to remove an electron from an isolated atom or molecule. Lesser the Ionization potential stronger is the reducing agent.

Oxidizing agent

$\rightarrow$ gains electrons and reduced.

Reducing agent

$\rightarrow$ loses electrons and oxidized.

Strongest Oxidizing agent

$\rightarrow$ D (IP - 700 and EA - 1035)

Strongest Reducing agent

$\rightarrow$ B (IP - 225 and EA - 700)

Option D is correct.

Pick the correct order of ionization potential values of elements.

(a) C < O < N < F < Ne

(b) He > Ne > Ar > Kr > Xe

(d) Be < Mg < Ca < Sr < Ba

(e) Na > Mg > Al > Si > P

0%
a, c and e
0%
b, c and d
0%
a, b and c
0%
b, d and c

Suppose a gas mixture of

$F, Cl, Br$ and

$I$ is irradiated with photons of frequency appropriate to ionize

$Cl.$ What ion(s) will be present in the mixture ?

0%
$F^+$ only
0%
$Cl^+$ only
0%
$F^+,\,Cl^+,\,Br^+$ only
0%
$Cl^+,\,Be^+,\,I^+$ only

Explanation

Variation in a group among the representative elements:

The ionization energy generally decreases in moving from top to bottom because the size increases due to the increase of the principal quantum number. On the other hand, the effective nuclear charge

$Z_{eff}$ for the outermost electron remains almost the same along the group.

As Ionization potential of

$Br$ and

$I$ is less than

$Cl$ so if the frequency is enough to ionize

$Cl,$ then I.E. of

$Br$ and

$I$ would be lesser than

$Cl,$ thus it can ionize them also.

So we will find all three cations in solution.

Which of the following statement(s) is(are) correct?

0%
The electron affinity for sulphur is more exothermic than that for oxygen
0%
Successive ionization energies of an atom always increase
0%
First ionization energy of $As$ is greater than that of $Se$
0%
Chlorine has larger atomic size as well as electron affinity than that of fluorine

Explanation

(A) The oxygen atom is smaller as compared to sulphur, so in oxygen, there is more inter-electronic repulsion than that of sulphur.

(B) Charge increases so size decreases and the number of electrons per proton also decreases, so valence shell electron gradually becomes more tightly held by the nucleus.

$As$ is

$ns^{2}np^{3}$ (half-filled stable configuration) and valence shell electron configuration of

$Se$ is

$ns^{2}np^{4}$ (partially filled less stable configuration).

(D) Down the group atomic size increases. Fluorine experiences more inter-electronic repulsion than chlorine, so fluorine has less electron affinity than that chlorine.

Consider the following statements.

$S_1$ : Fluorine does not form any polyhalide as it has low F - F bond energy

$S_2$ : The chlorine has the most negative electron gain enthalpy.

$S_3$ : The first ionization potential of N and 0 atoms are 14.6 and 13.6 eV respectively.

Which of the above statements are correct?

0%
$S_1, S_2$ and $S_3$
0%
$S_1$ and $S_2$
0%
$S_1$ and $S_3$
0%
$S_2$ and $S_3$

Explanation

$S_{1} :$ Fluorine does not form any polyhalide as other halogens because:- (1) It has maximum ionic character (2) It has low F-F bond energy (38.5 k cal mol–1) (3) Of the absence of d-orbitals in the valence shell of fluorine (4) It brings about maximum coordination number in other elements.

$S_{2} :$ The negative electron gain enthalpy of fluorine is less than that of chlorine. It is due to small size of fluorine atom. As a result, there are strong interelectronic repulsions in the relatively small 2p orbitals of fluorine and thus, the incoming electron does not experience much attraction.

$S_{3} :$ Consider the electronic configurations of Nitrogen and Oxygen.

N -

$1s^{2} 2s^{2} 2p^{3}$

O -

$1s^{2} 2s^{2} 2p^{4}$

We notice that if oxygen loses one electron, it will attain half-filled configuration. Hence it can lose an electron easily. Also, nitrogen is already in a half-filled configuration, and hence losing an electron will require more energy than that of oxygen.

Therefore The first ionization potential of N and 0 atoms are 14.6 and 13.6 eV respectively.

Hence , All are correct .

Therefore , option A is correct .

Ionization energy of

${He}^{+}$ is

$19.6 \times {10}^{-18} J {atom}^{-1}$ . The energy of the first stationary state

$\left(n=1\right)$ of

${Li}^{2+}$ is:

0%
$-2.2 \times {10}^{-15} J {atom}^{-1}$
0%
$8.82 \times {10}^{-17} J {atom}^{-1}$
0%
$4.41 \times {10}^{-16} J {atom}^{-1}$
0%
$-4.41 \times {10}^{-17} J {atom}^{-1}$

Explanation

$\dfrac{{I}_{{He}^{+}}}{{I}_{{Li}^{2+}}}=\dfrac{{Z}_{1}^{2}}{{Z}_{2}^{2}}$

$\dfrac{19.6 \times {10}^{-18}}{{I}_{{Li}^{2+}}}=\dfrac{4}{9}$

${I}_{{Li}^{2+}}=\dfrac{9}{4} \times 19.6 \times {10}^{-18} = 44.1 \times {10}^{-18}$

$=4.41 \times {10}^{-17} J {atom}^{-1}$

${E}_{{Li}^{2+}}=-4.41 \times {10}^{-17} J {atom}^{-1}$

Ionisation energy in group

$1A$ varies in the decreasing order as:

0%
$Li > Na > K > Cs$
0%
$Na > Li > K > Cs$
0%
$Li > Cs > K > Na$
0%
$K > Cs> Na > Li$

Explanation

Atomic size increases as we move from top to down in a group, therefore, the amount of energy required for ejection of an electron from atom decreases, i.e. ionisation energy decreases. Hence, the correct order of

$IE_{1}$ is

$Li > Na > K > Cs$ .

The ionisation enthalpy of

$He^{+}$ ion is

$19.60\times 10^{-18}J\ atom^{-1}$ . The ionisation enthalpy of

$Li^{2+}$ ion will be:

0%
$84.2\times 10^{-18}J atom^{-1}$
0%
$44.10\times 10^{-18}J atom^{-1}$
0%
$63.20\times 10^{-18}J atom^{-1}$
0%
$21.20\times 10^{-18}J atom^{-1}$
0%
$2.17\times 10^{-19}J atom^{-1}$

Explanation

Ionisation enthalpy of

$Li^{2+} = \dfrac {9}{4}\times ionisation\ enthalpy\ of\ He^{+}$

$= \dfrac {9}{4}\times 19.60\times 10^{-18}$

$= 44.1\times 10^{-18}J atom^{-1}$ .

Which has maximum ionisation potential?

0%
$N$
0%
$O$
0%
$O^{+}$
0%
$Na$

Explanation

Hint: The minimum amount of energy required to remove an electron from an isolated gaseous atom is called its ionization potential.

Correct Answer: Option C

Explanation:

Electronic configuration of the below elements are-

$N=1s^22s^22p^3$

$O=1s^22s^22p^4$

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

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

In option $C$ , oxygen has already released one electron and it shows a half-filled configuration. So, it will require higher energy than the other elements to remove an electron from it.

Final Answer: $O^+$ shows maximum ionization potential.

Considering the trend for ionization energy on the periodic table, which of the following elements has the highest ionization energy?

0%
Boron
0%
Oxygen
0%
Nitrogen
0%
Fluorine

Explanation

(D) F.
Ionisation Energy: This is the energy that is required to remove one electron from the valence shell.
Now, as the

$F$ has the tendency to gain one electron to complete its octet, hence it has the tendency to gain the electron not lose the electron.

Important point is that

$F$ does not form the positive oxidation state, as it is the most electronegative element.

Option D is correct.

Arrange the following elements in terms of increasing first ionization energy:

$Ga,\ Ba,\ Ru,\ F,\ N$

0%
$Ba < Ru < Ga < N < F$
0%
$Ga<Ba<Ru<F<N$
0%
$Ru<Ba<Ga<N<F$
0%
None of the above

Explanation

With the help of the periodic table, we can find the trend of the effective nuclear charge. It increases from left to right in a period and decreases down the group.
And more the effective nuclear charge, the higher will be ionisation energy.

Hence order will be as follows:

$Ba < Ru < Ga < N < F$

Option A is correct.

The table above shows the first five ionization energies for a second-period element. Which identifies the element and provides the best explanation?

	Ionization Energy $(kJ/mol)$
First	$801$
Second	$2,430$
Third	$3,660$
Fourth	$25,000$
Fifth	$32,820$

0%
Boron, because it has five electrons.
0%
Boron, because is has three valence electrons.
0%
Nitrogen, because it has five valence electrons.
0%
Nitrogen, because it has three electrons in a $2p$ sublevel.

Explanation

The fourth IE is very high compared to third IE, thus on removal of

$3$ electrons, the element attains inert gas configuration. Thus the element has

$3$ valence electrons. Hence among given options, Boron is correct because it has

$3$ valence electrons.

The most electropositive element is:

0%
$Na$
0%
$Al$
0%
$F$
0%
$K$

Explanation

The most electropositive element is

$K$ .
The most electronegative element is

$F$ .
On moving down the group, the electropositve character increases.
On moving from left to right in a period, the electropositve character decreases.

A swimmer coming out from a pool is covered with a film of water weighing about 18 g. Calculate the internal energy of vaporisation at

$100^0\ C$ .

[ ∆ vap H for water at 373 K = 40.66 kJ mol -1 ]

0%
$37.55 kJmol^{-1}$
0%
$3.67kJmol^{−1}$
0%
$30.67 kJmol^{-1}$
0%
$3.567kJmol^{−1}$
35.67kJmol−

Explanation

$H_2O(l) \xrightarrow{\text{vaporization}} H_2O(g)$

$m=18g$

Number of moles

$= \dfrac{18g}{18 g\ mol^{-1}} = 1\ mol$

$\Delta_{vap} U = \Delta_{vap} H^{\circleddash} - P\Delta V$

$=\Delta_{vap} H^{\circleddash} - \Delta ngRT$

$= 40.66-(1\times 8.314\times 10^{-3} \times 373)$

$=40.66 - 3.10$

$\Delta_{vap} U = 37.56kJ \ mol^{-1}$

Hence, the correct option is

$\text{A}$

The first (

$\Delta H_{1}$ ) and second (

$\Delta H_{2}$ ) ionisation enthalpies (in

$kJ mol^{-1}$ ) and the electron gain enthalpy (

$\Delta_{eg}H$ ) (in

$kJ mol^{-1}$ ) of the elements I, II, III, IV and V are given below:

Element	$\Delta_{1}H_{1}$	$\Delta_{1}H_{2}$	$\Delta_{eg}H$
I	520	7300	-60
II	419	3051	-48
II	1681	3374	-328
IV	1008	1846	-295
V	2372	5251	+48

The most reactive metal and the least reactive non-metal of these are respectively.

0%
I and V
0%
V and II
0%
II and V
0%
IV and V

Explanation

I represents

$Li$ , II represents

$K$ , III represents

$Br$ , IV represents

$I$ , V represents

$He$ . So, amongst these, II represents most reactive metal and V represents least reactive non-metal.

Also, II has the lowest value of first ionization enthalpy and V has positive electron gain enthalpy.

An element X has electronic configuration 2, 8,It forms ionic compounds with ions like nitrate, sulphate and phosphate. What will be the formulae of these compounds?

0%
${X}_{2}{NO}_{3}, {X}_{2}{SO}_{4},{X}_{2}{({PO}_{4})}_{3}$
0%
${X}_{2}{NO}_{3}, {X}_{}{SO}_{4},{X}_{2}{({PO}_{4})}_{3}$
0%
${X}_{}{NO}_{3}, {X}_{}{SO}_{4},{X}_{}{({PO}_{4})}_{3}$
0%
${X}{({NO}_{3})}_{2}, {X}_{}{SO}_{4},{X}_{3}{({PO}_{4})}_{2}$

If the configuration of

$Y^{+}$ &

$X^{-}$ is

$2s^{2}\ 2p^{6}$ then

0%
$IP$ of $X=IP$ of $Y$
0%
$EA$ of $X>EA$ of $Y$
0%
$IP$ of $X^{-}=IP$ of $Y^{+}$
0%
$IP$ of $X^{-}>IP$ of $Y^{+}$

In which of the following options order of arrangement does not match with the variation of property indicated against it?

0%
$Al^{3+} < Mg^{2+} < Na^{+} < F^{-}$ (increasing ionic size)
0%
$B < C < N < O$ (increasing first ionisation enthalpy)
0%
$I < Br < F < Cl$ (increasing electron gain enthalpy)
0%
$Li < Na < K < Rb$ (increasing metallic radius)

Explanation

Ionisation enthalpy of

$B$ is less than ionisation enthalpy of

$C$ but ionisation enthalpy of

$N$ is greater than ionisation enthalpy of

$O$ due to half-filled orbitals in

$N$ which have extra stability. Removal of an electron from half-filled, stable

$2p$ orbital requires more energy than removal of an electron from partially filled

$2p$ orbital of

$O$ .

Electron gain enthalpy of

$Cl$ is greater than electron gain enthalpy of

$F$ as

$Cl$ has bigger size and lesser electronic repulsion than

$F$ .

The decreasing order of the ionization potential of the following elements is:

0%
$Ne > Cl > P > S > Al > Mg$
0%
$Ne > Cl > P > S > Mg > Al$
0%
$Ne > Cl > S > P > Mg > Al$
0%
$Ne > Cl > S > P > Al > Mg$

Explanation

$\bf{Explanation:}$

Ionization potential is the amount of energy required to remove an outermost valence electron from the atom.
In moving down the group from top to bottom the ionization potential decreases and on moving from left to right in a period it is increases.
$Ne$ is the noble element, it has completely filled octet hence, it has the highest ionization potential.
Half-filled ( $P$ ) and completely filled configuration ( $Mg$ ) are the cause of the higher value of I.E.
The outer electronic configuration of $Mg$ and $Al$ are $\displaystyle 3s^2$ and $\displaystyle 3s^23p^1$ respectively.

The decreasing order of the ionization potential of the elements is:

$Ne>Cl>P>S>Mg>Al$ .

Hence, the correct answer is option $B$ .

Which of the following statement is correct?

0%
IP of $X_{(g)} >$ E.A. of $X_{(g)}$
0%
IP of $X_{(g)} >$ IP of $X^-{_{(g)}}$
0%
IP of $X^+_{(g)} >$ IP of $X_{(g)}$
0%
All are correct

Explanation

IP of

${ X }^{ + }\left( g \right) >IP\quad of\quad X\left( g \right)$ , Since Zeff of

${ X }^{ + }$ is more than

$X$ so more energy is needed to extract other

${ e }^{ - }$ .

Since

${ X }^{ - }$ (negatively charged species has more electron density, hence it can give easily electron so its IP is less than

$X$ )

$I\varepsilon$ is greater than

$\varepsilon .A$ since in

$\varepsilon .A$ of neutral species energy generally releases than in

$I\varepsilon$ which require more energy to extract an electron.

Generally, the first ionisation energy increases along a period, but there are some exceptions. From the given sets which is not an exception:

0%
N and O
0%
Na and Mg
0%
Mg and AI
0%
Be and B

Supposing the energy of the fourth shell for the hydrogen atom is -

$50$ a.u. (arbitrary unit). What would be its ionization potential?

0%
$50$
0%
$800$
0%
$15.4$
0%
$20.8$

$Cr$ ,

$21^{st}$ electron will enter in which orbital?

0%
$d_{xy}$
0%
$d_{yz}$
0%
$d_{zx}$
0%
Any of the above

The electronic configuration of an element is

$1s^22s^22p^63s^23p^3$ . The atomic number of the element, which is just below the above element in the periodic table is__

0%
32
0%
33
0%
35
0%
30

The

$I{ P }_{ 2 },I{ P }_{ 3 },I{ P }_{ 4 }\ and\quad I{ P }_{ 5 }$ of an element are 7.1, 14.3 , 34.5, 46.8, 162.2 eV respectively .The element is likely to be:

0%
Na
0%
Si
0%
F
0%
Ca

The second ionization energy is maximum for:

0%
boron
0%
beryllium
0%
magnesium
0%
aluminium

All alkali metal superoxides contain the

$[O_{2}]^-$ ion. They are:

0%
Paramagnetic
0%
Coloured compounds
0%
Oxidizing agents
0%
All of these

The enthalpy of formation of gaseous

$N_2O$ and

$NO$ at 298 K are 82 and 90

$kJ\ mol^{-1}$ . The enthalpy of reaction:

$N_2O(g)+\cfrac {1}{2}O_2\longrightarrow 2NO(g)$ is

0%
-8 kJ
0%
98kJ
0%
-74 kJ
0%
+8 kJ

In which pair do both atoms have one electron only in an "s" orbital in their ground states?

0%
$Ca, Sc$
0%
$Cu,Be$
0%
$H,He$
0%
$Li,Cr$

Which pair of elements has the same characteristic chemical properties?

0%
$Z=13, Z=22$
0%
$Z=3, Z=11$
0%
$Z=4, Z=24$
0%
$Z=2, Z=4$

In which of the following pairs, the first member has higher first ionization energy ?

0%
$N, O$
0%
$Cl, Ar$
0%
$Al, Ga$
0%
$B, Be$

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

0%
$K> Na> Li$
0%
$Ge> Si> C$
0%
$Be> Mg> Ca$
0%
$B> N> C$

Match the atomic numbers 19, 15, 8, 4 and 2 with each of the following:

0%
19: A metal of valency one.
0%
15: A solid non-metal of period 3
0%
8: A rare gas
0%
2: A gaseous element with valency 2.
0%
4: An element of group 2.

Two moles of

$N _ { 2 }$ and two moles of

$\mathrm { H } _ { 2 }$ are taken in a closed vessel of 5 litres capacity and suitable conditions are provided for the reaction. When the equilibrium is reached, it is found that a half mole of

$N _ { 2 }$ is used up. The equilibrium concentration of

$N H _ { 3 }$ is:

0%
0.3 M
0%
0.4 M
0%
0.2 M
0%
0.1 M

If each orbital can hold a maximum of three electrons, the number of elements in

$9 ^ { \text { th } }$ period of periodic table (long form) are

0%
$48$
0%
$162$
0%
$50$
0%
$75$

Which one of the following is a correct electronic composition of sodium?

0%
$2,8$
0%
$8,2,1$
0%
$2,1,8$
0%
$2,8,1$

Which members of each of the following pairs of atoms has higher first ionisation enthalpy and why?

0%
$Na : K$
0%
$Ar:Cl$
0%
$B: Be$
0%
$O: N$

Sodium forms

$Na^{ + }$ and not

$Na^{ 2+ }$ because :

0%
sodium contains only one electron in outermost shell.
0%
first ionization potential is small and the difference in first and second ionization potentials is very large.
0%
radius of $Na^{ + }$ is much smaller than of $Na^{2 + }$ .
0%
None of the above.

The first four ionization potentials (eV) are given for two elements

$X$ and

$Y$ .Identify them.

X	8.296	25.149	37.92	259.298
Y	5.318	47.29	71.65	98.88

0%
$B, Na$
0%
$Na, Be$
0%
$Be, B$
0%
$Na, Mg$

For the highlighted element in its

$X^{2-}$ form identify the CORRECT electronic configuration.

0%
$[Kr] 5s^2 4d^{10} 5p^4$
0%
$[Kr] 5s^2 5p^4$
0%
$[Kr] 5s^2 4d^{10} 5p^6$
0%
$[Kr] 5s^2 5p^6$

CBSE Questions for Class 11 Medical Chemistry Classification Of Elements And Periodicity In Properties Quiz 12 - 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 12 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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