MCQ Questions for Class 12 Engineering Chemistry The D-And F-Block Elements Quiz 10

Among

$d-$ block elements, the most abundant element belongs to :

0%
first transition series
0%
second transition series
0%
third transition series
0%
fourth transition series

Explanation

Most abundantly found transition elements are iron and titanium.

Iron is the most abundant element, by mass, in the Earth, constituting about 80% of the inner and outer cores of Earth. The molten outer core is about

$8000$ km in diameter, and the solid inner core is about 2400 km in diameter. Iron is plentiful because of its nuclear stability in stellar nuclear fusion reactions.

Titanium metal is a very durable metal for engineering applications because this metal is corrosion-resistant and also this metal is very strong and very light. It is

$40\%$ lighter than steel but as strong as high-strength steel. So titanium finds applications in things like aerospace.

Iron and Titanium belongs to

$1st$ transition series

$\mathbf{Hence\ the\ correct\ answer\ is\ (A).}$

The correct order of ionic radii of

$Ce, La, Pm$ and

$Yb$ in

$+3$ oxidation state is:

0%
$La^{3+}< Pm^{3+}< Ce^{3+}< Yb^{3+}$
0%
$La^{3+}< Ce^{3+}< Pm^{3+}< Yb^{3+}$
0%
$Yb^{3+}< Ce^{3+}< Pm^{3+}< La^{3+}$
0%
$Yb^{3+}< Pm^{3+}< Ce^{3+}< La^{3+}$

Explanation

Order of radii:

$Yb^{+3}< Pm^{+3} < Ce^{+3} < La^{+3}$

Lanthanide contraction is the greater-than-expected decrease in ionic radii of the elements in the lanthanide series from atomic number 57, lanthanum, to 71, lutetium.

Due to Lanthanide contraction,Ionic radii decreases continously.

Transition metals are often paramagnetic owing to

0%
Their high M.P. and B.P
0%
The presence of vacant orbitals
0%
The presence of one or more unpaired electrons in the system
0%
Their being less electropositive than the elements of groups I-A and II-A

Transition elements are coloured

0%
Due to small size
0%
Due to metallic nature
0%
Due to unpaired d - electrons
0%
All of these

Which of the following is not the characteristic of transition metals?

0%
They are all metals
0%
They show variable oxidation states which always differ by two units
0%
They are paramagnetic
0%
They easily form complex

Explanation

The

$d- block$ elements that are the elements from a group

$3-12$ are called transition elements. This is because their physical and chemical properties are intermediate between

$s-block$ and

$p-block$ properties. Hence these elements act as a bridge from

$s-block$ to

$p-block$ and hence they are known as transition

$\mathbf{Characteristic\ of\ transition\ metals}$

They have a large charge/radius ratio.

They are hard and have high densities.

They have high melting and boiling points.

They form compounds that are often paramagnetic.

They show variable oxidation states which differ by

$1$ unit

They form coloured ions and compounds.

They form compounds with profound catalytic activity.

They form stable complexes.

All the transition elements are metals; this is because of the number of electrons in the outermost shell. Transition metals are hard, malleable, and ductile due to the presence of strong metallic bonds.

$\mathbf{Hence\ the\ correct\ answer\ is\ (B)}$

The property which is not characteristic of transition metals, is:

0%
variable oxidation state
0%
tendency to form complex
0%
formation of coloured compounds
0%
natural radioactivity

Explanation

$\mathbf{Characteristic\ of\ transition\ metals}$

They have a large charge/radius ratio.
They are hard and have high densities.
They have high melting and boiling points.
They form compounds that are often paramagnetic.
They show variable oxidation states which differ by $1$ unit
They form coloured ions and compounds.
They form compounds with profound catalytic activity.
They form stable complexes.
All the transition elements are metals; this is because the number of electrons in the outermost shell is only. Transition metals are hard, malleable, and ductile due to the presence of strong metallic bonds.

A transition metal is one that forms one or more stable ions which have incompletely filled d orbitals.

$\mathbf{Radioactive\ elements }$

Radioactive elements are unstable isotopes that release subatomic particles or energy as they decay. Alpha decay releases two protons and two neutrons. It is potentially dangerous.

$\because$ The transition element are more stable they are not naturally radioactive.

$\mathbf{\because Transition\ metal\ (d-block elements)\ don't\ show\ radioactivity\ in\ general\ while\ these \ are}$

$\mathbf{ characterized\ to\ show\ variable\ oxidation\ states,\ form\ complexes\ and\ coloured\ compounds.}$

$\mathbf{Hence\ the\ correct\ answer\ is\ (D)}$

The radius of

$La^{3+}$ (atomic number of

$La = 57$ ) is

$1.06 \mathring{A}$ which one of the following given values will be closest to the radius of

$Lu^{3+}$ (atomic number of

$Lu = 71$ )?

0%
$1.40 \mathring{A}$
0%
$1.06 \mathring{A}$
0%
$0.85 \mathring{A}$
0%
$1.60 \mathring{A}$

Explanation

In lanthanides, there is fairly regular decrease in the size with increasing atomic numbers. So, possible radius for

$Lu^{3+}$ is

$0.85\ \overset{o}A.$

Ammonia solution dissolves fairly:

0%
$Hg_2Cl_2$
0%
$PbCl_2$
0%
$AgI$
0%
$Cu(OH)_2$

Explanation

Ammonia solution dissolves fairly in

$Cu(OH)_2.$ Copper(II) hydroxide reacts with a solution of ammonia to form a deep blue solution of tetramminecopper

$[Cu(NH_3)_4]^{2+}$ complex ion.

$Reaction:$

$Cu^{2+}+NH_3 \rightarrow [Cu(NH_3)_4]^{2+}$

Option D is correct.

Which one of the following ions is diamagnetic?

0%
$Cr^{2+}$
0%
$V^{2+}$
0%
$Sc^{3+}$
0%
$Ti^{3+}$

Which one of the following complexes is diamagnetic in nature?

0%
$[Cr(NH_3)_6]^{3+}$
0%
$[FeF_6]^{3-}$
0%
$[Co(NH_3)_6]^{3+}$
0%
$[Fe(H_2O)_6]^{2+}$

A metal ion from the first transition has a magnetic moment (calculated) of

$3.87\ B.M$ . How many unpaired electrons are expected to be present in the ion?

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Verdigris is:

0%
basic copper acetate
0%
basic lead acetate
0%
basic lead
0%
none

Explanation

Verdigris is basic copper acetate.

$(Verdigris -(CH_3COO)_2Cu.Cu.(OH)_2)$ is a blue-green powder used in green pigment and in dyes and also in the manufacture of insecticides and fungicides.

When excess of

$SnCl_2$ is added to a solution of

$HgCl_2$ , a white precipitate turning to grey is obtained. The grey colour is due to the formation of:

0%
$Hg_2Cl_2$
0%
$SnCl_4$
0%
$Sn$
0%
$Hg$

Which one of the following is paramagnetic in nature ?

0%
$Ni(CO)_4$
0%
$[Ni(CN)_4]^{2-}$
0%
$[NiCl_4]^{2-}$
0%
$[Co(NH_3)_6]^{3+}$

The colour of zinc sulphide is:

0%
yellow
0%
white
0%
brown
0%
black

Which of the following pairs of ions have same paramagnetic moment?

0%
$Mn^{2+}, Cu^{2+}$
0%
$Cu^{2+}, Ti^{3+}$
0%
$Ti^{4+}, Cu^{2+}$
0%
$Ti^{3+}, Ni^{2+}$

Explanation

$Cu^{2+}, Ti^{3+}$ have the same number of unpaired electrons thus they will have same paramagnetic moment.

The radius of

$La^{3+}$ is

$1.06\ \overset {o}{A}$ , which of the following given values will be closest to the radius of

$Lu^{3+}$ (At. no of

$Lu=71, La=57$ )?

0%
$1.6 \ {A}^{o}$
0%
$1.4\ {A}^{o}$
0%
$1.06\ {A}^{o}$
0%
$0.85 \ {A}^{o}$

Explanation

$La$ and

$Lu$ belongs to the lanthanoids (4f block elements of the inner transition elements) . Lanthanum is the

$1^{st}$ element and Luthentium is the last one. In lanthanoids the atomic an ionic radii decreases with the increase in atomic number.

The given options has on

$0.85 {A}^{o}$ satisfying the conditions .

Among

$d-$ block elements, the most abundant element belongs to the:

0%
first transition series
0%
second transition series
0%
third transition series
0%
fourth transition series

Which one of the following ion has a magnetic moment of

$5.93\ B.M$ ?

0%
$Mn^{2+}$
0%
$Fe^{2+}$
0%
$Cr^{2+}$
0%
$V^{3+}$
0%
$Cr^{3+}$

Which has maximum paramagnetic character ?

0%
$[Fe(CN)_6]^{4-}$
0%
$[Cu(H_2O)_4]^{2+}$
0%
$[Cu(NH_3)_4]^{2+}$
0%
$[Mn(H_2O)_6]^{2+}$

Explanation

$Mn^{+2}$ has

$d^5$ configuration with 5 unpaired electrons. Hence, it has the highest paramagnetic character.

Which of the following is not correct?

0%
$[Fe(CN)_6]^{3-}$ is weakly paramagnetic
0%
$[Fe(CN)_6]^{4-}$ is strongly paramagnetic
0%
$[Ni(CN)_4]^{2-}$ is diamagnetic
0%
$[NiCl_4]^{2-}$ is paramagnetic

Explanation

$[Fe(CN)_6]^{-4}$ is diamagnetic because it contains no unpaired electron.
1680578_1730922_ans_6e061e0e4dc14ece835dd167cf53ad8f.png

Which group contains coloured ions out of

$Cu^+\ Ti^{4+}\ Co^{2+}\ Fe^{2+}$

0%
$1,2,3,4$
0%
$3,4$
0%
$2,3$
0%
$1,2$

Explanation

Group

$3^{rd} \ and \ 4^{th}$ contains coloured ions

Which of the following pairs of transition metal ions are the stronger oxidising agents in aqueous solution?

0%
$V^{2+}$ and $Cr^{2+}$
0%
$Ti^{2+}$ and $Cr^{2+}$
0%
$Mn^{3+}$ and $Co^{3+}$
0%
$V^{2+}$ and $Fe^{2+}$
0%
$Ni^{2+}$ and $Fe^{2+}$

Which of the following statements is wrong for the transition elements ?

0%
Transition elements are placed from $3^{rd}$ to $6^{th}$ period
0%
Last electron enters in (n-1) d orbital
0%
Exhibits variable valency
0%
General electronic configuration is (n-1) $d^{1-10}$ $n s^{0-2}$

Explanation

Transition elements starts from

$4^{th}$ period as they have valence shell configuration

$\left ( n-1 \right )d^{1-10}ns^{0-2}$ .

Option A is correct.

Predict the magnetic nature of

$A$ and

$B$ .

0%
both are diamagnetic
0%
$A$ is diamagnetic and $B$ is paramagnetic with one unpaired electrons
0%
$A$ is diamagnetic and $B$ is paramagnetic with two unpaired electrons
0%
both are paramagnetic

Explanation

$[Ni(CN)_4]^{2−}$

The oxidation number of

$Ni$ is

$+2$ .

Atomic number = 28

$Ni = [Ar]3d^84s^2$

$Ni^{2+}=[Ar]3d^8$

There are no unpaired electrons, so the complex is diamagnetic.

$[NiCl_4]^{2−}$

The oxidation number of

$Ni$ is

$+2$ .

Atomic number = 28

$Ni = [Ar]3d^84s^2$

$Ni^{2+}=[Ar]3d^8$

Chloride is a weak field ligand, no pairing.

$sp^3$ hybridization (tetrahedral)

There are two unpaired electrons, so the complex is paramagnetic.

The complex forming tendency of a transition metal depends upon:

0%
Availability of a number of vacant $d-$ orbitals
0%
high ionisation energy
0%
small size of cation or high charge density
0%
variable oxidation states

The correct order of magnetic moments amongst following complex ions is:

0%
$[MnCl_4]^{2-} > [Fe(CN)_6]^{4-} > [CoCl_4]^{2-}$
0%
$[Fe(CN)_6]^{4-} > [MnCl_4]^{2-} > [CoCl_4]^{2-}$
0%
$[MnCl_4]^{2-} > [CoCl_4]^{2-} > [Fe(CN)_6]^{4-}$
0%
$[Fe(CN)_6]^{4-} > [CoCl_4]^{2-} > [MnCl_4]^{2-}$

Explanation

Experimental evidence of magnetic measurements supports the theory of high- and low-spin complexes. Remember that molecules such as

$O_2$ that contain unpaired electrons are paramagnetic. When an electron in an atom or ion is unpaired, the magnetic moment due to its spin makes the entire atom or ion paramagnetic.

$Mn$ is in a

$+2$ oxidation state having configuration of

$[Ar]3d^5$ . It has

$5$ unpaired electrons.

$\mu = \sqrt{n(n+2)}= \sqrt{(5(5+2))}=5.9BM$

The magnetic moment of

$[MnC1_4]^{2−}$ is

$5.9 BM$ .

$Co$ is in

$+2$ oxidation state having the configuration of

$[Ar]3d^7$ . It has

$3$ unpaired electrons.

$\mu = \sqrt{n(n+2)}= \sqrt{(3(3+2))}=3.9BM$

Magnetic moment of

$[CoC1_4]^{2−}$ is

$3.9 BM$ .

$Fe$ is in

$+2$ oxidation state having the configuration of

$[Ar]3d^6$ . In this complex

$CN^−$ ion is a strong ligand and all electrons are paired and there are no unpaired electrons.

$\mu = \sqrt{n(n+2)} = \sqrt{(0(0+2))} =0BM$

Magnetic moment of

$[FeCN_6]^{4-}$ is

$0.0 BM.$

$\mathbf{\therefore [MnCl4]^{2−}>[CoCl4]^{2−}>[Fe(CN)6]^{4−}}$

$\mathbf{Hence\ the\ correct\ answer\ is\ option\ (C)}$

In the long form of periodic table, all the non-metals and metalloids are placed with :

0%
$s$ -block
0%
$p$ -block
0%
$d$ -block
0%
$f$ -block

Explanation

In the long form of the periodic table, all the non-metals and metalloids are placed

$p-block$ .

$\mathbf{Hence\ the\ correct\ answer\ is\ option\ (B)}$

1562288_1733896_ans_630bc7e4d0714c85a329a4bee3f54435.jpg

Which one of the following complexes is not diamagnetic?

0%
$[Fe(CN)_6]^{4-}$
0%
$[Ti(H_2O)_6]^{3+}$
0%
$[Pt(CN)_4]^{2-}$
0%
$[Co(NH_3)_6]^{3+}$

Explanation

Complexes having unpaired electrons are

$Paramagnetic$ .

Complexes having paired electrons are

$Dimagnetic$ .

(i)

$[Fe(CN)_6]^{4-}$

Fe oxidation state = +2 i.e.

$([Ar]3d^6)$

$CN^-$ is a strong ligand and all 3d electrons in Fe gets paired.

$[Fe(CN)_6]^{4-}$ is low spin complex and

$d^2sp^3$ hydridized.

Hence it is dimagnetic.

(ii)

$[Ti(H_2O)_6]^{3+}$

Ti oxidation state = +3 i.e.

$([Ar]3d^1)$

$H_2O$ is a weak field ligand and one unpaired electron is present in 3d orbital of Ti.

$[Ti(H_2O)_6]^{3+}$ is high spin complex and

$d^2sp^3$ hydridized.

Hence it is paramagnetic.

(iii)

$[Pt(CN)_4]^{2-}$

Pt oxidation state = +2 i.e.

$([Xe]5d^{10})$

$CN^-$ is a strong ligand and all 5d electrons in Pt gets paired.

$[Pt(CN)_4]^{2-}$ is low spin complex and

$d^2sp$ hydridized.

Hence it is dimagnetic.

(iv)

$[Co(NH_3)_6]^{3+}$

Co oxidation state = +3 i.e.

$([Ar]3d^6)$

$NH_3$ is a strong ligand and all 3d electrons in Co gets paired.

$[Co(NH_3)_6]^{3+}$ is low spin complex and

$d^2sp^3$ hydridized.

Hence it is dimagnetic.

$[Ti(H_2O)_6]^{3+}$ is paramagnetic.

Therefore, the correct option is

$(B)$ .

The correct increasing order of paramagnetism of

$(i)\ MnSO_4. 4H_2O$

$(ii)\ FeSO_4. 7H_2O$

$(iii)\ NiSO_4. 6H_2O$

$(iv)\ CuSO_4. 5H_2O$ is:

0%
$(i) < (ii) < (iii) < (iv)$
0%
$(iv) < (iii) < (i) < (i)$
0%
$(iii) < (iv) < (ii) < (i)$
0%
$(iii) < (iv) < (i) < (ii)$

Explanation

Paramagnetism depends on number of unpaired electrons. More the unpaired electrons, more will be paramagnetism.

(i)

$MnSO_4.4H_2O$

Mn oxidation state = +2 i.e

$[Ar]3d^5$

Number of unpaired electrons = 5

(ii)

$FeSO_4.7H_2O$

Fe oxidation state = +2 i.e.

$[Ar]3d^6$

Number of unpaired electrons = 4

(iii)

$NiSO_4.6H_2O$

Ni oxidation state = +2 i.e.

$[Ar]3d^8$

Number of unpaired electrons = 3

(iv)

$CuSO_4.5H_2O$

Cu oxidation state = +2 i.e.

$[Ar]3d^9$

Number of unpaired electrons = 1

Therefore, Increasing order of paramagnetism is

$(iv)<(iii)<(ii)<(i)$

Hence correct option is (B).

Transition metals form complexes because of:

0%
small cation size
0%
vacant $d-$ orbitals
0%
large iconic charge
0%
all of these

The ionic radius of

$_{57}La^{3+}$ is 1.06

$\overset oA$ Which one the following give values will be closest to the ionic radius of

$_{71}Lu^{3+}$ ?

0%
1.06 $\overset oA$
0%
1.56 $\overset oA$
0%
0.85 $\overset oA$
0%
1.35 $\overset oA$

Which ine of the following statement about lanthanides is false?

0%
Lanthanides are sepearated from one another by ion exchange method
0%
The ionic radii of trivalent lanthanides steadily increases with increase in atomic number
0%
All lanthanides are highly dense metals
0%
Most typical oxidation state of lanthanides is +3

The ions from among the following which are colourless are

(i)

$Ti^{4+}$ (ii)

$Cu^{+}$ (iii)

$Co^{3+}$ (iv)

$Fe^{2+}$

0%
i and ii only
0%
i, ii and iii
0%
iii and iv
0%
ii and iii

Explanation

$Ti^{4+} \Rightarrow [Ar] 3d^04s^0$ , No unpaired

$e^-$

$Cu^{+} \Rightarrow [Ar] 3d^{10}4s^0$ , No unpaired

$e^-$

$Co^{3+} \Rightarrow [Ar] 3d^64s^0$ , 4 unpaired

$e^-$

$Fe^{2+} \Rightarrow [Ar] 3d^64s^0$ , 4 unpaired

$e^-$

$Ti^{4+}$ and

$Cu^{+}$ are colourless.
Option A is correct.

Which of the following function as catalysts for oxidation of

$SO_2$ to

$SO_3$ in industrial process for production of

$H_2SO_4$ ?

0%
$TiO_2$
0%
$PbO_2$
0%
$Pt$
0%
$V_2O_5$

Among the following, the species that is both paramagnetic and coloured is :

0%
$\left [ MnO_{4} \right ]^{2-}$
0%
$\left [ TiCl_{6} \right ]^{2-}$
0%
$\left [ VO_{4} \right ]^{2-}$
0%
$CrO_{2}Cl_{2}$

Which ionic radius of

${_{57}{La^{3+}}}$ is 1.06A. which one of the following values will be closest to ionic radius of

${_{71}{Lu^{3+}}}$ . ?

0%
1.06A
0%
1.56A
0%
0.85A
0%
1.35A

Explanation

Ionic radius of

${Lu{3+}}$ is smaller than that of

${La{3+}}$ because of lanthanide contraction across the period.

Out of all the known elements, the percentage of transitional elements is approximately

0%
$30\%$
0%
$50\%$
0%
$60\%$
0%
$75\%$

Explanation

No. transition element = 33

Total known element = 105

Percentage of transitional elements

$=\dfrac { \text{Transition element }}{\text{ total element} } \times 100$

$= \dfrac {33 }{105} \times 100 = 31\approx 30$ %

Atomic number of Cr and Fe are 24 and 26 respectively. Which of the following is paramagnetic with the spin of electron?

0%
$[ Cr(NH_3)_6]^{3+}$
0%
$[Fe(CO)_5]$
0%
$[ Fe (CN)_6]^{4-}$
0%
$[ Cr(CO)_6 ]$

Explanation

Since,

$NH_3$ is a weak field ligand.

Hence, it will not pair up electrons, whereas

$CO$ ,

$CN^-$ being strong field ligand will not pair up electrons.

No. of unpaired electrons in

$[ Cr(NH_3)_6]^{3+}$ ,

$[Fe(CO)_5]$ ,

$[ Fe (CN)_6]^{4-}$ &

$[ Cr(CO)_6 ]$ are 3, 0, 0 & 0 respectively.

Hence, the only paramagnetic complex among the given is

$[ Cr(NH_3)_6]^{3+}$ .

Hence, Option "A" is the correct answer.

Which of the following pair of elements cannot form an alloy ?

0%
$Zn - Hg$
0%
$Fe - Hg$
0%
$Hg - Cr$
0%
$Na - Hg$

Iron loses magnetic property at

0%
Melting point
0%
1000K
0%
Curie point
0%
Boiling point

The term platforming is

0%
Platinum painting
0%
Flat sheet of platinum
0%
Platinum manufacturing
0%
Platinum used as a catalyst

Arrange

$Ce^{3+} , La^{3+} ,Pm^{3+}$ and

$Yb^{3+}$ in increasing order of their ionic radii

0%
$Yb^{3+} < Pm^{3+} < Ce^{3+} < La^{3+}$
0%
$Ce^{3+} < Yb^{3+} < Pm^{3+} < La^{3+}$
0%
$Yb^{3+} < Pm^{3+} < La^{3+} < Ce^{3+}$
0%
$Pm^{3+} < La^{3+} < Ce^{3+} < Yb^{3+}$

Explanation

Due to Lanthanoid contraction order will be:

$Yb^{3+} < Pm^{3+} < Ce^{3+} < La^{3+} .$

Colour of cobalt chloride solution is

0%
pink
0%
black
0%
colourless
0%
green

Explanation

Colour of anhydrous

$CoCl_2$ is blue and when it comes in contact with moisture then it turns pink.

Which of the following statement(s) is(are) correct with reference to the ferrous and ferric ions?

0%
$Fe^{3+}$ gives brown colour with potassium ferricyanide
0%
$Fe^{2+}$ gives blue precipitate with potassium ferricyanide
0%
$Fe^{3+}$ gives red colour with potassium thiocyanate
0%
$Fe^{2+}$ gives brown colour with ammonium thiocyanate

Alizarin gives a red colour by mordanting it with the sulphate of a metal. The metal ion involved is

0%
$Cr^{3+}$
0%
$Fe^{3+}$
0%
$Al^{3+}$
0%
$Sn^{2+}$

Explanation

"Alizarin" is a typical mordant dye gives different colours depending on the metal ion used , for example with

$Al^{3+}$ , alizarin gives a rose red colour.
Co-ordination compound of alizarin with

$Al^{3+}$ as follow
1679613_1765524_ans_8c4e26ea5aab4414b19d8795f506965d.png

$CdI_2$ is pink in colour, the

$CdCl_2$ will be '________' coloured.

0%
yellow
0%
red
0%
blue
0%
cannot be predicted

Explanation

$CdI_2$ Pink colour is due to polarization of

$I^-$ by

$Cd^{2+}$ has pseudonoble gas configuration. In

$CdCl_2$ , extent of polarization of

$Cl^-$ by

$Cd^{2+}$ is relatively less, hence specific colour cannot be predicted.

Which metal absorbs hydrogen?

0%
$Al$
0%
$Fe$
0%
$Pd$
0%
$K$

In the dichromate anion

$(Cr_{2}O_{7}^{2-}),$

0%
all $Cr - O$ bonds are equivalent
0%
$6 \,Cr - O$ bonds are equivalent
0%
$3 \,Cr - O$ bonds are equivalent
0%
no bonds in $Cr_{2}O_{7}^{2-}$ are equivalent.

Which of the following statements is wrong?

0%
$Ti^{4+}$ and $Ag^{+}$ are repelled by magnetic field
0%
$Mn^{2+}$ shows maximum magnetic character among the first transition series.
0%
$Fe^{2+}$ is more stable than $Mn^{2+}$ towards oxidation to $+3$ state.
0%
$Cr$ in $Cr_{2}O_{7}^{2-}$ ion involves $sp^{3}d^{2}$ hybridisation

CBSE Questions for Class 12 Engineering Chemistry The D-And F-Block Elements Quiz 10 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry The D-And F-Block Elements Quiz 10 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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