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Momic vadiue a tt 'S defined asthe distance between nucleus of the atom & tho electyoy cloud of the outermost shell, Claseiliratioy of ahomic vadiu« l | cat . Covalent 2. Metalie Védius Yadiu 3. Vaanderwaal'¢ Yadius l. Covalent vadive To —_ Covalent vadive may be defined as half of the dictance botweon ne centres of the nuclei of two similar alos bonded by a Single Covalent bond. — example: Distance between two #-atomns ty Hy moleculeg has been found 10 be eaual tO Oma 4. ES Therefore Covaleit radius Covalent vadive of ye atom = 1 y 0.44% Ye= dap Z 7 a = 0.394 L How 40 calculate coval "ke molecule — UW) If AEN (Ka & Xp) ts Very Small dh-a = Yat ey aD) If AEN (Xa BxQ) 1s Nery large /Jarg¢ dA-= Yat Yy — 0.09 (xs-Xq) KAS EN OL & Xp = EN ofs ent yadius of and & in AG 2% Metattic yadiue . inter nuclear Metallic vadius is defined as hal ‘ i fallic latte. distance between two Adyacont atoms in Ho to Cxamble: the dictawe behweon two adjacent cu atoms in solid cu ie 2-sCh. once Metallic vad, . o Yadus of cy = [ax 2esJh = |°28A H Ym = da-s u da-a 2— 2= 3. Nauder wagl'e Yadius Sheer waal's vadiug _ His defined as half of the huclei of two hon- bonded nelghbouring oom bolonaing two adsacent molecules of DN len ent in the solid state . OES) Exaimpla: the dictaneo between nuclei MM ! of tuo Odgacent ‘es’ atoms of two Ay viele te adsacent ‘uw Molecules in Solid ctate ig SCA. So, n= t = da_, Vander waalls yadiue Of at’ Nader waa) 2 akon is 3-¢4° = sa 2 a 9. She metallic vadine of au atom ig always larger than its Covalent yadius — why > : Ans—> Because metallic bond is weaker than Covalent bond wt example tm (Ol Na)= (ace and that of y ‘nt State) — Isqye / S*elege (in Hs “oP. ?. Covalent Yadiug of an atom je Shorter than ifs Vandeywaal'¢ Yadiug — why » ANS > Because the formation of covalent bond Involves OuerapPing of atomic orbitals for which i distance between Ter hue leay +wo Covalently bonded atoms the inter nuclear distance is less than between two hou- bow, * key concept: ded atome. ‘Covalent < Vmetallie Z Yvander waa} =e ce defined as the distance between “he nuclew ¢ of tho jong aad +he pot upto whieh the nucleug exerts its force on the electron cloud ofthe ion. Q. the yvadiug of cation is always smaller than that of the Neutral atom— explain. Ans—> with the vemoval of electrons trom an atom, Zeff incveases and Vadius detreases. For example radius of Na= (544° 8 joe vadiue of Nak ise 0.958", 9. the radius of Chlovine atom is 0.594° but ionic radius of Cl ts |-844 — explain. AN$—> el + —7 ie With the addition of ons extra electron, Zep of el becomes lesser than Zeft of cd and thus size of A becomes 9reater than du atom- 9. the vadius of amon is always greater than that of the neutral atom — oxblain. Ay Due 40 addition of eledrons Ze of Anion decreases and lonic Yadius increases. For example Y= 0.998 but v7 = 184A. Key concebt —— Teselechonie species ave those which have tame numbey Of electons. wet ae E 24-7 Gyt® Ar, KT cl, = p= ete. for these sheciee the Size (ionic vadius) decreases with an incase *f nuclear charge i. atomic number, -lonisation botentiat or mea ental The am Clech dunt energy Yequved +0 vemove the-most loosely bound element rom Nalence shell of an iso lated gaseous atom of the Ing § lowect energy state +o can atom into Fascous cation. , gy state, +0 convert +hz 1S Called tonisation Potential or jonisation enthalpy 40 ¢ H 2, is the minimum amount of energy vequired h oMNert any Jascous akom Cm) into its gaseous Cation Mt en +the 10NSSahon €nthalpy of M is Ty: = M(a)+ 1, —> M*(g)+@ t Aet Le. This is an endothermic protess, $0 AH= +ve. 9. 4, L2nd ionisation enthalpy) is higher than 1, [dst ionisation enthalpy) — ecblain. . Ang—y Let ‘ty ts the minimum amoun} of energy vequiyed +0 Convert any gasequs atom in its ground state into Saseous ion Mt then - M(q)+ 1,—> Mt(9)+ € St 7 the minimum amount of energy required to Comey M+(9) into Mt*(4) then M+(4)4+ 1, —> w*4(4) 42 Tt is more difficult to yemoye an electron from a cation than from a neutyal atom. Itence £55, , Similarly 39 £, and $9 on ie. dtydrjg------ 2. Factors attecting ionisation enthalpy Cl) Size of the atom: T.€. cc | Size of atom cu) Etfective nucleay chavge (Zep): LE of Zeft qi) _Seveening effect or Schielding effed: We Know Zey= B2B- ellechve nuclear charge, Seteening censfant 2e > Hective Nuclear and <-> cteening ool 4y Case of tt, it has ctable hal f Cled p- orbitals Where aS ‘o' atom has one electron more than at of ‘w' atom. $0, t€-4 of ‘0’ is lesser than that Of N’.. Now aftey yemoval of one @ fom SO' atom, tf attains stable 29° con figuyalion. 60, Ot has least tenclency +o loose the stability of b-ovbital, hence and L€ of O '$ mote than that of N. Q. Woblec Gases ov inert Jases have very high value of L.E- explain, . Ansg—> Flectvonic configuration of outermost Shell of wert Gases 1s ns np> which is except onally stable electronic configuration, $0, a large amount of energy, is vequired +o Convert @ Woulval inary gas atom into its 10N, SO, they Wane high values of 4-f. Q. Alkali metals have low Values of 1-£.- explain. A-> Gen. electronic ConfiguvaHon of alkali metal is Cn-s)s' Cn) ns! ETwo outermost Shells} So, loss of One electron trom their outermost chal) bying & about stable electronic Configuration of inert gases, Shevetore \ess amount of energy is vequired 40 Convert a neutral alkali metal atom into its ton. As a tesulk alkali metals have low Values of LE. 
