chem mcqs........

[intel1993]

i have problem in m/j 2006 p1
q31 n 13...................

 

[kevin_ram19]

Hi,

13. In which pair is the radius of the second atom greater than that of the first atom?
A .Na, Mg
B .Sr, Ca
C .P, N
D .Cl, Br

Answer: D.
Explanation:
-Going down a group in periodic table of elements, number of electrons increases, number of shells increases, thus atomic size increases.
-going across a period, although the number of electrons increases, they fill the same shell. But, nuclear attraction toward the electron shells increases as the protons number increases. Thus, the atomic size decreases across a period.
-Mg is smaller than Na, so A wrong
-Sr, is larger than Ca, so B is wrong
-P is larger than N, so c is wrong
-Br is larger than Cl----so D is correct

 

[lee91]

qs 13,
across period, the size of atom decreases, because the outermost electron increases, so the attractive force between the nucleus and the electrons increases. Shielding effect is not necessary to be considered because acorss period the number of inner shell electrons remain the same, so shielding effect would be more or less the same.

down the group, the size of atom increase because inner shell electrons increases, so shielding effect increases. Here, the nuclear charge also increase but the increase in shielding effect outweigh the icnrease in nuclear charge, so you only have to consider the effect of shielding effect, sounds confusing huh..

q 31,
1, down the group, the number of electrons increases, so the orbital overlap will decrease due to greater repulsion of electrons when no of electrons increase, so if the orbital overlap decrease, (or even if it's the same,) still the bond length will increase down the group.
2, of course atomic radius decrease down the group, as what i've said above.
3, nuclear charge increases down the group, this statement is true, but it is not explaining why the bond length increases, imagine, if nuclear charge increase, then the attractive froce between nucleus and outermost electrons will be stronger so the atom will be smaller and hence, then bondlength between the two atoms will decrease not increase.

Hope this clarify, and i'm not quite sure with q 31 choice 1, i believe the explanation is correct, it's the only explanation i can think of.

 

[intel1993]

hey sry it was mj 2005 not 06...

so can u help me now....
mj 2005 q13 n 31.......

 

[lee91]

mj05
q13,
A and D definitely out, it is too far from aluminium, we only have to consider B and C.
to consider the electronegative, we need to look at the atomic radius, as atomic radius increase, the electronegative decrease.
so, magnesium has a bigger radius than aluminium, so has lower electronegativity,
but, beryllium has smaller radius than magnesium, so has higher electronegativty.
If the electronegativity of aluminium is 1.5, then electronegativity of magnesium will be 1.2, and then electronegativity of beryllium will be larger than 1.2, probably 1.5, so answer B.

q31,
1, definitely AlCl3 is trigonal planar, there is no lonepair left.
3, definitely wrong, PH3 there is a lone pair at P.
2, CH3+, the positive meaning the wrong molecule is lacking of one electron, imagine if the electron is from C, there will be 3 outermost electron left, so the 3 will be shared with hydrogen to form single covalent bond, hence, there is no lone pair left on C as well, so a trigonal planar.

 

[intel1993]

k thank u very much....
i better uinderstand the electronegativity concept nw.......