D and F block Flashcards
ELECTRONIC CONFIG
- Sc 21
- Ti 22
- V 23
- Cr24
- Mn 25
- Fe 26
- Co 27
- Ni 28
- Cu 29
- Zn 30
- [Ar]3d1 4s2
- [Ar]3d2 4s2
- [Ar]3d3 4s2
- [Ar]3d5 4s1
- [Ar]3d5 4s2
- [Ar]3d6 4s2
- [Ar]3d7 4s2
- [Ar]3d8 4s2
- [Ar]3d10 4s1
- [Ar]3d10 4s2
D block elements?
why are they called transition elements?
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.
How do u explain the catalytic property of transition metals
YEAH THAT LIST ONLY
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.
which metal is used as catalyst in
1. contact process
2. haber’s process
3. catalytic hydrogenation
4. iodide and persulphate ions
- vanadium(V) oxide
- finely divided iron
- nickel
- iron (III)
how does transition metal work as catalyst
Eais lowering
- 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)
- also bcz transition metal ions can change their oxid states , they become more effective as catalysts
WHY Sc3+, cu+, Zn2+, Ti4+ colourless
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.
why transition metals form coloured ions
** 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
why the centre metal atom of Complez compound A TRANSITION METAL
only time when short, negative and vacant is fine
BECAUSE:
1. SMALL SIZE
2. high effective nuclear charge or high ionc charge
3. availability of d orbitals or vaccant d orbital
a single unpaired e- has magnetic moment?
1.73 BM (bohr magneton)
calculate the magnetic moment of a divalent ion of atomic no. 25
- d5 config
- √5(5+2)= 5.92BM
Silver atom has completely filled d orbitals (4d10) in its ground state. How can you say that it is a transition element?
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.
Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?
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.
how are INTERSTITIAL compounds formed
AND their phy and chem characters
when small atoms of H, C, OR N are trapped in the CRYSTAL lattice of METALS
- HIGH MP
- HARD
- RETAIN METALLIC CONDUCTIVITY
- CHEMICALLY INERT
ABOUT what percent should the metallic radii be to form ALLOYS
give examples of FERROUS ALLOYS
15%
- Cr
- V
- W
- Mo
- Mn
BRASS
BRONZE
- cu and zn
- cu and tin
OXOCATIONS stabilises:
1. V(V)
2. V(IV)
3. Ti(IV)
into?
- VO2^+
- VO^2+
- TiO^2+
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
- grp no
- Sc
- Fe2O3
WHAT happens to ionic and covalent character as oxidation number** INCREASE**
IONIC CHARACTER DECREASES (LESS BASIC)
COVALENT CHARACTER INCREASES (MORE ACIDIC)
CLASSIFY AS ACIDIC, BASIC OR AMPHOTERIC:
1. Mn2O7 (green oil)
2. CrO3
3. V2O5
4. CrO
5. Cr2O3
6. V2O3
7. V2O4
- acidic
- acidic
- amphoteric
- basic
- amphoteric
- basic
- less basic
what happens when V2O5 reacts with acidic and alkali medium
V2O5 & CrO3 have low MP
- on acidic medium : VO4^+
- on alkali medium : VO4^3-
which metals have HIGH VOLATILITY AND SOFT
no. of unpaired e- increases interatomic bonding increases and HARDNESS increases
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
why in 4d and 5d series metals have greater enthalpies of atomisation
notLANTHANOIDCONTRACTIONTSKSKKS
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).
why does atomic radii decreases along the series (3d)
AGAINNOT LANTHANOID CONTRACTIONTSSKSKSK
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**
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?
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.
why does atomic radii decreases along the series IN 4d and 5d
YEAHSAYIT
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)
what happens to DENSITY from titanium to copper
- atomic mass increases and metallic radius decreases
- Results in INCREASE in DENSITY
ACCOUNT FOR THE FOLLOWING
- Irregular trend in first ionisation enthalpy in 3d series
- increase of ionisation enthalpy is slight from Sc to Zn 🛡🛡
- increase in values of SECOND ionisation enthalpy🍀
- general rend in I.E breaks for Mn2+ and Fe3+ in second and third I.E
- removal of one e- alters the RELATIVE ENERGIES of 4s and 3d
- 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
- effective nuclear charge increase beacuse on d e- cant shield another d e- as d- orbitals differ in direction
- beacause both have d5 config
three terms responsible for I.E
- attraction of e- towards nucleus
- repulsion between e-
- EXCHANGE ENERGY
WHAT IS RESPONSIBLE FOR STABILISATION OF ENERGY STATE
how it is related to ionisation
Exchange energy
- proportional to possible pairs of parallel spin in degenerate orbitals
- loss of exchange enrgy increases stability thus ionisation becomes more difficult
at which config there is no loss of exchange energy
d6
ACCOUNT FOR THE FOLLOWING
- I.E FOR Mn+ is lower than Cr+
- I.E for Fe 2+is lower than Mn2+
- 3rd I.E of Fe is lower than Mn
- 2nd I.E for Cu and Cr is unusually high
- 2nd I.E for Zn is low
- high 3rd I.E for Mn and Zn
- difficult to obtain oxi state higher than +2 for Ni, Cu and Zn
- Mn+ - 3d5 4s1and Cr+ - d5 It is exactly half-filled and hence its ionization enthalpy is higher than Mn+.
- 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
- 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
- they have stable d5 and d10 config
5.as ionisation causes removal of one 4s e- which resultls i stable d10 config
- removal of e- causes configuration to change from stable d5 an d10 to unstable configs
- 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.
most common lowest oxidation state
in d block
+2
name a transition element which doesn’ exhibit variable oxidation state
scandium
* bcz Sc+3 has noble gas config and is stable
account for the following
- why titanium shows +4 oxidation state
- why vanadiums shows +5 oxi state
they acquire NOBLE GAS CONGFIG
WRITE CONFIG AND SHOW
Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?
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.
WHY does lesser no. of oxidation states STEM from extreme ends of the transition metal series
there are either too few e- to lose or share
Acount for the following:trends in M2+/M S.E.P
- E0 value more -ve for Mn
- E0value more -ve for Zn
- E0 valure more -ve for Ni
- E0 value more +ve for Cu
- stability of the half filled d sub shell in Mn 2+
- stability of the completely filled d subshell in Zn 2+
- E0 related to the highes negative ΔHhyd enthalpy of hydration
- high enthalpy of atomisation and low enthalpy of hydration
Account for the trends in M3+/M2+ S.E.P
- high value for zn
- low value for Sc
- high value for Mn
- low value for Fe
- low value for V
- removal of e- from stabe d10 config of Zn2+
- stability of Sc3+ which has a noble gas config
- Mn2+ has d5 which is particularly stable
- extra stability of fe3+ (d5)
- stability of V2+ (t2g)
Account for the foollowing:
- THE ability of fluorine to stabilise the highes oxidation state in CoF3
- Vf3 and CrF3
- the ability of oxygen to stabilise these high oxidation state exceed that of Flourine
- higher lattice energy
- higher bond enthalpy
- the ability of oxygen to form multiple bonds
Why is the highest oxidation state of a metal exhibited by its fluoride and oxide only?
Both fluorine and oxygen have very high electronegativity values. They can oxidise the metals to the highest oxidation state.
Explain why Cu+ ion is not stable in aqueous solutions?but Cu2+ is.
so what happens to Cu(I) {unstable} in aqueous soln
- 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
Which is a stronger reducing agent Cr2+ or Fe2+ and why?
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
Explain the irregualrity of the first row transition metals in their E0 values
due to irregular ionisation enthalpies and also sublimation enthalpies which are much less for mangenes and vanadium
why is Eo value for Mn3+/Mn2+couple more +ve than Cr3+/Cr2+ or Fe3+/Fe2+
much larger third ionisation energy of Mn where the need to change d5 to d4 config
potassium di chromate
ore:
preparation
oxidising reactions
chromite ore (FeCr2O4)
- fusion of chromite ore with sodium or pottasium carbonate in free access of air
- the yellow sodium chromate is filteres and acidified with sulphuric acid
- 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+
potassium permanganate
ore:
preparation:
Mno2 (pyrolusite)
- fusion of Mno2 with alkali metal hydroxide and oxidising agent KNO3
- dark green K2MnO4 is produced which disproportionates in a neutral and acidic soln to give Permanganate
commercial and lab preparation of KMnO4
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+
decomposition of KMnO4
physical properties of KMnO4
colour of K2MnO4
tetrahedral: the pi bonding by overlap of p orbi- O2 & s orb-Mn
- 2KMnO4 -> K2MnO4 + MnO2 + O2
- diamagnetic with temp dependent weak Paramagnetism and dark purple colour
- paramagnetism due to unpaired e- and dark green colour
permangante reduction
1. e-
2. 4H+ + 3e-
3. 8H+ +5e-
- MnO42-
- MnO2 + 2H2O
- Mn2+ + 4H2O
Lanthanide contraction
cause
consequence
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
WHY ZIRCONIUM AND HOFNIUM HAS SIMILAR ATOMIC RADII
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)
COMMON oxidation state in LANTHANOIDES
+3
BUT ALSO SHOWS +2 AND +4
WHY IS CERIUM+4 A GOOD OXIDISING AGENT
BECAUSE IT LOSE ONE E- TO FORM Ce+3
f0 ANDf14 CONFIG ARE COLOURLESS AND diamagnetic like La+3 and Lu+3
due to absence of unpaired e-
which f block elements have +4 oxi state in their oxides
- Nd
- Pr
- Tn
- Dy
f0 and f14 are diamagnetic and colourless
due to absence of unpaired electron
steel hard f block element
samarium 1623K
mischmetal consist of
95% lanthanide metal
5% fe
traces of S, C, Ca. and Al
- used in Mg based alloy to produce bullets, shell and lighter flint
lanthanoid known to exhibit +4 OXIDATION STATE
Ce
