Hi i have another question relating to the atomic radius of
transition metals.
for element 1-20, going across a group, screening effect
remains constant and the only factor affecting atomic
radius is due to the increasing nuclear charge and thus increasing
attraction between valence electrons and positive
nucleus.
However, such a trend is not observed for transition metals;
their atomic radius stays relatively unchanged even with increasing
proton no. This is because electrons are added into the inner 3d
subshells and not 4s subshell, and hence the screening
effect increases. (is it correct to say this?) The
increasing screening effect offsets the the increase in nuclear
charge and thus attraction between 4s electrons and nucleus is
relatively same. Hence, atomic radius remains relatively
unchanged.
Are my explanations correct or is there a better way of phrasing
it?
Your answer is correct.
JC students commonly have difficulty with this topic, because it's
just like the classic accountant joke (the boss is hiring and
interviewing a physicist, a mathematician, a IT guy and an
accountant; the boss asks, "what's 2 + 2?" The physicist, a
mathematician, and IT guy each give a similar expected answer, with
technical variants depending on their different backgrounds. When
it comes to the accountant, the accountant replies, "What do you
want it to be, boss?". The accountant got the job).
For the qn, "Why do d block elements have smaller atomic radii
compared to s block elements?", students are expected to state,
"3d orbitals are poor shielders!"
In contrast, for the qn, "Why do the atomic radii and 1st
ionization energy of d block elements remain relatively invariant
across the period?", students are expected to state, "3d
orbitals are effective shielders!"
The key to understanding this apparent contradiction, is of course
(Einstein would be proud) : Relativity.
For the same electron shell, when compared to the 3s and
3p orbitals, 3d orbitals are poor shielders because of their shape.
Hence d block elements have smaller atomic radii compared to s
block elements.
At the same time, 3d orbitals (being part of the less diffused 3rd
electron shell) provide more effective shielding compared to 4s
orbitals (being part of the more diffused 4th electron shell).
Accordingly, as we go across the period, because the additional
electron is added into the 3d orbitals, which provide relatively
more effective shielding compared to the 4s orbitals, and
consequently the slight increase in relatively effective shielding
by the additional 3d electron, partially offsets or counters the
increase in nuclear charge, resulting in the relative invariance of
the atomic radii and 1st ionization energy of the d block elements
across the period.
Anyway, that's just my own phrasing. Your phrasing is already
perfectly fine, and will gain you full credit by Cambridge in the A
levels.
Edited : Oh, in case some students are still wondering, "If for
the 2nd period, the shielding effect remains constant, then why
doesn't the shielding effect also remain constant for the d block
elements?"
The reason is : even for period 2, the shielding effect isn't
absolutely constant. It does increase slightly across the period,
when more electrons are added. But because electrons are added to
the 2nd electron shell (which is the outermost valence
shell), hence shielding effect from the 2nd electron shell is
relatively negligible (shielding effect is mostly effective from
inner shell electrons). In contrast, for the d block
elements, the additional electron is added into the 3d orbital
(which is part of an inner electron shell), which
does provide significantly effective additional
shielding effect for the outer 4s electrons, which are the
electrons involved in determining atomic radii and 1st ionization
energy.
