here consider the oscillation of centre of mass

driving force is mgsin@

acceleration = gsin@

a=g@=g(x/l)

here g and l are constant

therefore a is proportional to x

so the motion is SH

t=2pi underroot of displacement/acc.n

=2pi underroot of x/gx/l

rate if helped

naaaaaaaa.......

as we go down the no. of f and d electrons will increase and so will the nuclear charge......

u've misread my previous one

for such type of bodies the general formula is
A1X1 - A2X2
divided by A1 - A2
similarly for y coordinate

i hope so this formula will help ya

as the d or f electrons are situated very far frm each other

they give the nucleus to strengthen their hold on the s-electrons

and as s electrons show high penetrating effect ie they prefer to live near the nucleus

the hold of nucleus on 6s electrons is considerable...

plz rate if helped

inert pair effect is nothing put the inability of s electrons to participate in the reaction.

s electrons are not able to participate bcoz of strong hold of nucleus (high penetration effect) or por shielding effect of electrons

therefore

+3 state stability will be in the order

Bi>Sb>As

and vice-versa for +5 stability

more big the atom more poor the shielding effect...

 

 

catenation is nothing but the property of atom to form giant structures

rate me if dis infrmation is useful

basic nature of elements is directly proprtnl to metallic nature

coz more the metallic nature the easier is to loose the electron..

this criteria well satisfies the defination of a lewis base

which readily donates electron.

crct me if rong

f directly proportional to R to the powr -5/2
also F=m omega square R
F is proportional to omega square R
so omega square R is proportional to R to the powr -5/2
omega square is proportional to R to the powr -5/2
omega square is proportional to R to the powr -7/2
solve it and get the ansr which is 3

plz rate if helpful