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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2004

MCQ (Single Correct Answer)
Which one of the following ions has the highest value of ionic radius?
A
O2-
B
B3+
C
Li+
D
F-

Explanation

From a neutral atom where a electron is removed then it become cation and when a electron is added to the atom then it becomes anion.

For a atom $$x$$ when it loose a electron then it becomes cation $$x - {e^ - } \to {x^ + }$$

Let atomic no. of $$x$$ is $$z$$

then no. of proton in $${x^ + } = z$$

and no. of electron in $${x^ + } = z - 1$$

So the ratio $${z \over e}$$ as no. of electrons decreases. And when no of electron decreases then per electron attraction increases from nucleus and radius of cation deceases.

Similarly when atom $$x$$ becomes $${x^ - }$$ by adding a electron $$\left( {x + {e^ - } \to {x^ - }} \right)$$ then no. of proton in $${x^ - } = z$$ and no. of electron in $${x^ - } = z + 1.$$

Now the ratio of $${z \over e}\,\,$$ in $${x^ - }$$ decreases as no. of electron increases. As electron increases in $${x^ - }$$ then attraction from nucleus decreases on per electron and the distance between electron and nucleus increases so the radius also increase in anion.

All $$4$$ ions are from second period and for $${O^{2 - }}\,\,$$ and $${F^ - }$$ both have $$1s,$$ $$2s$$ and $$2p$$ shell and for $$L{i^ + }\,\,$$ and $${B^{3 + }}$$ both have $$1s$$ shell.

So, size of $${O^{2 - }}\,\,\,$$ and $$\,\,\,{F^ - }\,\,\,$$ are more then $$L{i^ + }\,\,\,\,$$

and $$\,\,\,\,{B^{3 + }}$$ and among $${O^{2 - }}\,\,\,$$ and $$\,\,\,\,{F^ - }$$

For $$\,\,\,$$ $${O^{2 - }},\,\,{z \over e} = {8 \over {10}} = 0.8$$

For $$\,\,\,$$ $${F^ - },\,\,\,{z \over e} = {9 \over {10}} = 0.9$$

as for $${O^{2 - }},\,\,{z \over e}\,\,\,$$ is less so per electron attraction from nucleus decreases and radius increases more than $${F^ - }.$$
2

AIEEE 2003

MCQ (Single Correct Answer)
The atomic numbers of Vanadium (V), Chromium (cr), Manganese (Mn) and Iron (Fe), respectively, $$23,24,25$$ and $$26$$. Which one of these may be expected to have the higher second ionization enthalpy?
A
Cr
B
Mn
C
Fe
D
V

Explanation

Ionization enthalpy is the energy required to remove the electron from the outer most orbit.

Electronic configuration :

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,V\left( {23} \right) = \left[ {Ar} \right]\,\,\,3{d^3}\,\,\,4{S^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,Cr\left( {24} \right) = \left[ {Ar} \right]\,\,\,3{d^5}\,\,\,4{S^1} \cr & \,\,\,\,\,\,\,\,\,\,\,Mn\left( {25} \right) = \left[ {Ar} \right]\,\,\,3{d^5}\,\,\,4{S^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,Fe\left( {26} \right) = \left[ {Ar} \right]\,\,\,3{d^6}\,\,\,4{S^2} \cr} $$

After first ionization enthalpy the electronic configuration will be,
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,V\left( {23} \right) = \left[ {Ar} \right]\,\,\,3{d^3}\,\,\,4{S^1} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,Cr\left( {24} \right) = \left[ {Ar} \right]\,\,\,3{d^5} \cr & \,\,\,\,\,\,\,\,\,\,\,Mn\left( {25} \right) = \left[ {Ar} \right]\,\,\,3{d^5}\,\,\,4{S^1} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,Fe\left( {26} \right) = \left[ {Ar} \right]\,\,\,3{d^6}\,\,\,4{S^1} \cr} $$

Now in 2nd ionization enthalpy one more electron will be removed. And after second ionization enthalpy electronic configuration will be,
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,V\left( {23} \right) = \left[ {Ar} \right]\,\,\,3{d^3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,Cr\left( {24} \right) = \left[ {Ar} \right]\,\,\,3{d^4} \cr & \,\,\,\,\,\,\,\,\,\,\,Mn\left( {25} \right) = \left[ {Ar} \right]\,\,\,3{d^5} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,Fe\left( {26} \right) = \left[ {Ar} \right]\,\,\,3{d^6} \cr} $$

For $$V(23), Mn (25)$$ and $$Fe (26)$$ electron is removed from $$4S$$ orbital but for $$Cr (24)$$ electron is removed from $$3d$$ orbital.

We know distance of $$4S\,\, > \,\,3d$$ from the nucleus. So, the attraction on the electrons of $$45$$ shell is less compared to $$3d$$ shell by the nucleus. So, the removed of electron will be easier for the electrons of $$4S$$ shell than $$3d$$ shell.

So, for $$Cr(24)$$ second ionization is more than other as electron is removed from $$3d$$ shell.

And for $$Cr(24)$$ outer most shell $$''3d''$$ was half filled after $$1st$$ ionization enthalpy, so $$3d$$ shell was stable. So removal of electron will be difficult from stable shell. That is why also $$2nd$$ ionization enthalpy of $$Cr(24)$$ is higher compare to other given atoms.
3

AIEEE 2003

MCQ (Single Correct Answer)
The reduction in atomic size with increase in atomic number is a characteristic of elements of
A
$$d$$ - block
B
$$f$$ - block
C
Radioactive series
D
High atomic masses.

Explanation

This can't be $$f$$ - block elements as $$f$$ block means only lanthanides and actinides but we know lanthanide contraction happens on the elements before lantanidine elements also. So, radius also decrease for those element when increase the atomic number.

Radioactive Series starts from atomic number $$84$$ ( Polonium ) and affer that all are radioactive. But we know before $$84$$ atomic number also the atomic size decrease with increase in atomic number. So, only radioactive series don't show this characteristic.

So, correct answer should be high atomic mass.
4

AIEEE 2003

MCQ (Single Correct Answer)
Which one of the following is an amphoteric oxide?
A
Na2O
B
SO2
C
B2O3
D
ZnO

Explanation

These elements are called amphoteric which reacts with both acid and base.

Following elements are amphoteric

$$\left( 1 \right)\,\,\,\,Zn$$

$$\left( 2 \right)\,\,\,\,Be$$

$$\left( 3 \right)\,\,\,\,Al$$

$$\left( 4 \right)\,\,\,\,B$$

$$\left( 5 \right)\,\,\,\,Cr$$

$$\left( 6 \right)\,\,\,\,Ga$$

$$\left( 7 \right)\,\,\,\,Pb$$

$$\left( 8 \right)\,\,\,\,Sn$$

Here $$Zn,Be,Al,Pb\,\,\,$$ and $$\,$$ $$Sn$$ are mostly amphoteric.

$$B, Cr$$ and $$Ga$$ are less amphoteric.

Questions Asked from Periodic Table & Periodicity

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