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Chemistry > D and F block > Flashcards

D and F block Flashcards

1
Q

ELECTRONIC CONFIG

  1. Sc 21
  2. Ti 22
  3. V 23
  4. Cr24
  5. Mn 25
  6. Fe 26
  7. Co 27
  8. Ni 28
  9. Cu 29
  10. Zn 30
A
  1. [Ar]3d1 4s2
  2. [Ar]3d2 4s2
  3. [Ar]3d3 4s2
  4. [Ar]3d5 4s1
  5. [Ar]3d5 4s2
  6. [Ar]3d6 4s2
  7. [Ar]3d7 4s2
  8. [Ar]3d8 4s2
  9. [Ar]3d10 4s1
  10. [Ar]3d10 4s2
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2
Q

D block elements?
why are they called transition elements?

A

metals which have incomplete d subshell either in **NEUTRAL ATOM **or in THEIR IONS

  • they exhibit transitional behaviour between s-block and p-block elements.
  • Their properties are transitional between highly reactive metallic elements of s-block which are ionic in nature and the elements of p-block which are covalent in nature.
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3
Q

How do u explain the catalytic property of transition metals

YEAH THAT LIST ONLY

A

Transition metals show catalytic behaviour mainly due to the following reasons:
* The presence of vacant d orbitals
* They have the ability to exhibit variable oxi st.
* They have a tendency to form complex compounds.

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4
Q

which metal is used as catalyst in
1. contact process
2. haber’s process
3. catalytic hydrogenation
4. iodide and persulphate ions

A
  1. vanadium(V) oxide
  2. finely divided iron
  3. nickel
  4. iron (III)
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5
Q

how does transition metal work as catalyst

Eais lowering

A
  1. catalysts at solid surface involve the formation of bonds btwn reactant molecules and atoms of the surface of the catalyst
  • this has the effect of increasing the conc of the reactants at the catalyst surface and also weakening the bonds in the reacting molecules (Eais lowering)
  1. also bcz transition metal ions can change their oxid states , they become more effective as catalysts
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6
Q

WHY Sc3+, cu+, Zn2+, Ti4+ colourless

A

Sc3+, Ti4+, Cu+ and Zn2+ have** either empty or filled 3d-orbital, **which means there is no presence of the unpaired d-electron. Therefore, they are colourless.

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7
Q

why transition metals form coloured ions

A

** due to the presence of unpaired electron**
* when e- from LOW ENERGY d orbital is excited to HIGHER ENERGY d orbital , the ENERGY OF EXCITATION corresponds to th frequency of light absorbed

  • This frequency lies in the VISIBLE REGION
  • the colour observed correponds to COMPLEMENTARY COLOUR OF THE LIGHT ABSORBED

the μ of light is determined by the NATURE OF LIGAND

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8
Q

why the centre metal atom of Complez compound A TRANSITION METAL

only time when short, negative and vacant is fine

A

BECAUSE:
1. SMALL SIZE
2. high effective nuclear charge or high ionc charge
3. availability of d orbitals or vaccant d orbital

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9
Q

a single unpaired e- has magnetic moment?

A

1.73 BM (bohr magneton)

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10
Q

calculate the magnetic moment of a divalent ion of atomic no. 25

A
  • d5 config
  • √5(5+2)= 5.92BM
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11
Q

Silver atom has completely filled d orbitals (4d10) in its ground state. How can you say that it is a transition element?

A

The outer electronic configuration of Ag (Z=47) is 4d105s1. It shows+1 and + 2 O.S. (in AgO and AgF2). i.e, d-subshell is incompletely filled. Hence, it is a transition element.

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12
Q

Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?

A

Manganese (Z = 25) shows maximum number of O.S. This is because its outer EC is 3d54s2. As 3d and 4s are close in energy, it has maximum number of e-1 s to loose or share. Hence, it shows O.S. from +2 to +7 which is the maximum number.

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13
Q

how are INTERSTITIAL compounds formed
AND their phy and chem characters

A

when small atoms of H, C, OR N are trapped in the CRYSTAL lattice of METALS

  1. HIGH MP
  2. HARD
  3. RETAIN METALLIC CONDUCTIVITY
  4. CHEMICALLY INERT
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14
Q

ABOUT what percent should the metallic radii be to form ALLOYS

give examples of FERROUS ALLOYS

A

15%

  • Cr
  • V
  • W
  • Mo
  • Mn
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15
Q

BRASS
BRONZE

A
  1. cu and zn
  2. cu and tin
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16
Q

OXOCATIONS stabilises:
1. V(V)
2. V(IV)
3. Ti(IV)
into?

A
  1. VO2^+
  2. VO^2+
  3. TiO^2+
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17
Q

FILL IN THE BLANKS
1. highest oxi state coincide with____________ from Sc2O3 to Mn2O7
2. expect _____________ all metals form MO oxides which are ionic
3. beyond grp 7 no higher oxides of Fe above_______ are known

A
  1. grp no
  2. Sc
  3. Fe2O3
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18
Q

WHAT happens to ionic and covalent character as oxidation number** INCREASE**

A

IONIC CHARACTER DECREASES (LESS BASIC)
COVALENT CHARACTER INCREASES (MORE ACIDIC)

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19
Q

CLASSIFY AS ACIDIC, BASIC OR AMPHOTERIC:
1. Mn2O7 (green oil)
2. CrO3
3. V2O5
4. CrO
5. Cr2O3
6. V2O3
7. V2O4

A
  1. acidic
  2. acidic
  3. amphoteric
  4. basic
  5. amphoteric
  6. basic
  7. less basic
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20
Q

what happens when V2O5 reacts with acidic and alkali medium

V2O5 & CrO3 have low MP

A
  1. on acidic medium : VO4^+
  2. on alkali medium : VO4^3-
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21
Q

which metals have HIGH VOLATILITY AND SOFT

no. of unpaired e- increases interatomic bonding increases and HARDNESS increases

A

ZN, Cd. Hg (complete filled d subshell)
*transition metals are hard and have low volatility

metal with high enthaply of atomisation (high BP) = noble in rnx

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22
Q

why in 4d and 5d series metals have greater enthalpies of atomisation

notLANTHANOIDCONTRACTIONTSKSKKS

A

Thus, the valence electrons are less tightly held and hence can form metal-metal bond more frequently. (That is why* melting points* of 4d and 5d series as well as* enthalpies of atomisation* are higher than those of 3d series).

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23
Q

why does atomic radii decreases along the series (3d)

AGAINNOT LANTHANOID CONTRACTIONTSSKSKSK

A

as new ** e- enters** d orbital, ** nuclear charge increases**
shielding effect of d orbitals isn’t effective and HENCE net electrostatic attraction ** Between nuclear charge and Outermost e- increases and atomic and ionic radii DECREASES**

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24
Q

In the series Sc(Z = 21) to (Z = 30), the enthalpy of atomisation of zinc is the lowest i.e., 126 kJ mol-1. Why?

A

no electrons from the 3d-orbitals are involved in the formation of metallic bonds since all the orbitals are filled. metallic bonds are quite weak in zinc and it has therefore, lowest enthalpy of atomisation in the 3d series.

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25
Q

why does atomic radii decreases along the series IN 4d and 5d

YEAHSAYIT

A

intervention of 4f and 5f orbitals which are poorer shielders than d orbitals lead to increase in nuclear charge and net electrostatic attraction bten nuclear charge and e- increases and THEREFORE ATOMIC RADIII DECREASES (lanthanoid contraction)

26
Q

what happens to DENSITY from titanium to copper

A
  1. atomic mass increases and metallic radius decreases
  2. Results in INCREASE in DENSITY
27
Q

ACCOUNT FOR THE FOLLOWING

  1. Irregular trend in first ionisation enthalpy in 3d series
  2. increase of ionisation enthalpy is slight from Sc to Zn 🛡🛡
  3. increase in values of SECOND ionisation enthalpy🍀
  4. general rend in I.E breaks for Mn2+ and Fe3+ in second and third I.E
A
  1. removal of one e- alters the RELATIVE ENERGIES of 4s and 3d
  2. nuclear charge increases along series but e- are added to the orbital of inner subshell i.e, 3d. 3d shield 4s electron from increasing nuclear charge somewhat better than outer most e-. thus atomic radii decrease less rapidly and I.E increase slightly
  3. effective nuclear charge increase beacuse on d e- cant shield another d e- as d- orbitals differ in direction
  4. beacause both have d5 config
28
Q

three terms responsible for I.E

A
  1. attraction of e- towards nucleus
  2. repulsion between e-
  3. EXCHANGE ENERGY
29
Q

WHAT IS RESPONSIBLE FOR STABILISATION OF ENERGY STATE
how it is related to ionisation

A

Exchange energy

  • proportional to possible pairs of parallel spin in degenerate orbitals
  • loss of exchange enrgy increases stability thus ionisation becomes more difficult
30
Q

at which config there is no loss of exchange energy

A

d6

31
Q

ACCOUNT FOR THE FOLLOWING

  1. I.E FOR Mn+ is lower than Cr+
  2. I.E for Fe 2+is lower than Mn2+
  3. 3rd I.E of Fe is lower than Mn
  4. 2nd I.E for Cu and Cr is unusually high
  5. 2nd I.E for Zn is low
  6. high 3rd I.E for Mn and Zn
  7. difficult to obtain oxi state higher than +2 for Ni, Cu and Zn
A
  1. Mn+ - 3d5 4s1and Cr+ - d5 It is exactly half-filled and hence its ionization enthalpy is higher than Mn+.
  2. Fe2+ has d6 config and Mn2+ has d5. the removal of electron requires more energy as it has to go from a stable electronic configuration (3d⁵) to unstable electronic configuration (3d⁴). Hence, Fe+² has lower ionization enthalpy than that of Mn+²
  3. Fe2+ has d6 config and Mn2+ has d5. the removal of electron requires more energy as it has to go from a stable electronic configuration (3d⁵) to unstable electronic configuration (3d⁴). Hence, Fe+² has lower ionization enthalpy than that of Mn+²
  4. they have stable d5 and d10 config

5.as ionisation causes removal of one 4s e- which resultls i stable d10 config

  1. removal of e- causes configuration to change from stable d5 an d10 to unstable configs
  2. bcz of high values of** 3rd ionisation enthalpies** or After removal of two electrons they will have fully filled d-orbitals which acquires extra stability,so it is difficult to obtain oxidation state greater than 2 for Cu,Niand Zn.
32
Q

most common lowest oxidation state
in d block

A

+2

33
Q

name a transition element which doesn’ exhibit variable oxidation state

A

scandium
* bcz Sc+3 has noble gas config and is stable

34
Q

account for the following

  • why titanium shows +4 oxidation state
  • why vanadiums shows +5 oxi state
A

they acquire NOBLE GAS CONGFIG

WRITE CONFIG AND SHOW

35
Q

Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?

A

Manganese (Z = 25) shows maximum number of O.S. This is because its outer EC is 3d54s2. As 3d and 4s are close in energy,** it has maximum number of e-1 s to loose or share**. Hence, it shows O.S. from +2 to +7 which is the maximum number.

36
Q

WHY does lesser no. of oxidation states STEM from extreme ends of the transition metal series

A

there are either too few e- to lose or share

37
Q

Acount for the following:trends in M2+/M S.E.P

  1. E0 value more -ve for Mn
  2. E0value more -ve for Zn
  3. E0 valure more -ve for Ni
  4. E0 value more +ve for Cu
A
  1. stability of the half filled d sub shell in Mn 2+
  2. stability of the completely filled d subshell in Zn 2+
  3. E0 related to the highes negative ΔHhyd enthalpy of hydration
  4. high enthalpy of atomisation and low enthalpy of hydration
38
Q

Account for the trends in M3+/M2+ S.E.P

  1. high value for zn
  2. low value for Sc
  3. high value for Mn
  4. low value for Fe
  5. low value for V
A
  1. removal of e- from stabe d10 config of Zn2+
  2. stability of Sc3+ which has a noble gas config
  3. Mn2+ has d5 which is particularly stable
  4. extra stability of fe3+ (d5)
  5. stability of V2+ (t2g)
39
Q

Account for the foollowing:

  1. THE ability of fluorine to stabilise the highes oxidation state in CoF3
  2. Vf3 and CrF3
  3. the ability of oxygen to stabilise these high oxidation state exceed that of Flourine
A
  1. higher lattice energy
  2. higher bond enthalpy
  3. the ability of oxygen to form multiple bonds
40
Q

Why is the highest oxidation state of a metal exhibited by its fluoride and oxide only?

A

Both fluorine and oxygen have very high electronegativity values. They can oxidise the metals to the highest oxidation state.

41
Q

Explain why Cu+ ion is not stable in aqueous solutions?but Cu2+ is.
so what happens to Cu(I) {unstable} in aqueous soln

A
  • This is becuase ΔhydH of Cu2+(aq) is much higher than that of Cu+(aq) and hence it compensates for the *2nd IE of Cu. *
  • Thus, many Cu(I) compounds are unstable in aqueous solution and undergo disproportionation as follows :
    2 Cu+ —–> Cu2+ + Cu
42
Q

Which is a stronger reducing agent Cr2+ or Fe2+ and why?

A

Cr2+ is a stronger reducing agent than Fe2+. This is because E°(Cr3+/Cr2+) is negative (- 0.41V) whereas E°(Fe3+/Fe2+) is positive (+ 0.77 V).

It is because Cr2+ loses electron to become Cr3+ which is more stable due to half filled t2g orbitals

43
Q

Explain the irregualrity of the first row transition metals in their E0 values

A

due to irregular ionisation enthalpies and also sublimation enthalpies which are much less for mangenes and vanadium

44
Q

why is Eo value for Mn3+/Mn2+couple more +ve than Cr3+/Cr2+ or Fe3+/Fe2+

A

much larger third ionisation energy of Mn where the need to change d5 to d4 config

45
Q

potassium di chromate
ore:
preparation
oxidising reactions

A

chromite ore (FeCr2O4)

  1. fusion of chromite ore with sodium or pottasium carbonate in free access of air
  2. the yellow sodium chromate is filteres and acidified with sulphuric acid
  3. sodium dichromate is treated with KCl and orange crytals of Potassium dichromate CRystallise out
  • iodide - iodine
  • sulphide to sulphur
  • tin(II) to tin (IV)
  • Fe(II) to Fe(III)

Cr2O7^2- + 14H+ + 6Fe2+ —> 2Cr3+ + 7H2O + 6Fe3+

46
Q

potassium permanganate
ore:
preparation:

A

Mno2 (pyrolusite)

  1. fusion of Mno2 with alkali metal hydroxide and oxidising agent KNO3
  2. dark green K2MnO4 is produced which disproportionates in a neutral and acidic soln to give Permanganate
47
Q

commercial and lab preparation of KMnO4

A

commercial:
1. alkaline oxidative fusion of MnO2
MnO2 _________fused with KOH ,oxidised with air or KNO3____ MnO42-
2. electrolytic oxidation of MnO42-
Mno42- ___________electrolytic oxidation in alkaline soln____ MnO4-

lab preparation:
1. mangenese(II) ion salt is oxidised by peroxodisulphate to permanganate
2Mn2+ + 5S2O82- + 8H2O = 2MnO4- + 10SO42- + 16H+

48
Q

decomposition of KMnO4
physical properties of KMnO4
colour of K2MnO4

tetrahedral: the pi bonding by overlap of p orbi- O2 & s orb-Mn

A
  1. 2KMnO4 -> K2MnO4 + MnO2 + O2
  2. diamagnetic with temp dependent weak Paramagnetism and dark purple colour
  3. paramagnetism due to unpaired e- and dark green colour
49
Q

permangante reduction
1. e-
2. 4H+ + 3e-
3. 8H+ +5e-

A
  1. MnO42-
  2. MnO2 + 2H2O
  3. Mn2+ + 4H2O
50
Q

Lanthanide contraction
cause
consequence

A

Lanthanide contraction is the steady decrease in atomic and ionic size of lanthanoids as their atomic number increases.

the poor shielding of 4f orbital from the increasing effectivenuclear charge

difficulty in separation
4d and 5d series have almost similar atomic series
ionic character decreases and covalent character decreases
and basic strenght decreases
electronegativity increases
formation of complex compound

51
Q

WHY ZIRCONIUM AND HOFNIUM HAS SIMILAR ATOMIC RADII

A

DUE TO LANTHANOID CONTRACTION

Due to poor shielding of f-subshell,
the attraction of nucleus on outer shell electrons increases
and thus the size of Hf decreases and size of Zr and Hf becomes almost the same.

ZR(160PM) AND Hf (159)

52
Q

COMMON oxidation state in LANTHANOIDES

A

+3
BUT ALSO SHOWS +2 AND +4

53
Q

WHY IS CERIUM+4 A GOOD OXIDISING AGENT

A

BECAUSE IT LOSE ONE E- TO FORM Ce+3

54
Q

f0 ANDf14 CONFIG ARE COLOURLESS AND diamagnetic like La+3 and Lu+3

A

due to absence of unpaired e-

55
Q

which f block elements have +4 oxi state in their oxides

A
  1. Nd
  2. Pr
  3. Tn
  4. Dy
56
Q

f0 and f14 are diamagnetic and colourless

A

due to absence of unpaired electron

57
Q

steel hard f block element

A

samarium 1623K

58
Q

mischmetal consist of

A

95% lanthanide metal
5% fe
traces of S, C, Ca. and Al

  • used in Mg based alloy to produce bullets, shell and lighter flint
59
Q

lanthanoid known to exhibit +4 OXIDATION STATE

A

Ce

60
Q

actinoide contraction is greater than lanthanoid contraction

A

due to poor shielding by 5f-electrons in actinoids than that by 4f-electrons in the lanthanoids.

Chemistry (9 decks)

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