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SIDDANNA MARPALLE. Govt. Neelambika PU College Basavakalyan Dist : Bidar. Q: 1. The lowest first ionisation energy would be associated with which of the following structures? a] 1s 2 2s 2 2p 6 3s 1 b] 1s 2 2s 2 2p 5 c] 1s 2 2s 2 2p 6 d] 1s 2 2s 2 2p 6 3s 2 3p 2.
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SIDDANNA MARPALLE Govt. Neelambika PU College Basavakalyan Dist : Bidar
Q: 1. The lowest first ionisation energy would be associated with which of the following structures? a] 1s2 2s2 2p6 3s1 b] 1s2 2s2 2p5 c] 1s2 2s2 2p6 d] 1s2 2s2 2p6 3s2 3p2 Ans : a] 1s22s2 2p6 3s1
Q: 2. Which of the following combinations contains only isoelectronic species? a] N3-, O2-, Cl- b] F-, Ar, S2-, Cl- c] P3-,S2-, Cl- d] N3-, F-, O2-, Ar Ans: c] P3-,S2- ,Cl-
Q: 3. Fluorine has unexpectedly lower electronic affinity than that of chlorine due to.
a] Big size of fluorine, which results in increase in the electron-electron repulsion 2P subshell b] Small size of fluorine, which results in increase in the electron-electron repulsion 2p-subshell. c] Decrease in electron-electron repulsion d] No electron-electron repulsion in 2p- subshell Ans: b] Small size of fluorine, which results in increase in the electron-electron repulsion 2p-subshell
Q: 4. The most electronegative element in the periodic table is. a] Nitrogen b] Oxygen c] Chlorine d] Fluorine Ans: d] Fluorine
Q: 5. Which one of the following has the smallest radius? a] Cl- b] S2- c] K+ d] Ca2+ Ans: d] Ca2+
Q: 6. Atomic size increase as we move down a group because. a] effective nuclear charge increases b] Atomic mass increases. c] Additive elements are accommodated in new electron level. d] Atomic number increases. Ans: c] Additive elements are accommodated in new electron level.
Q: 7. Electrons which are characterised by the outer shell configuration ns1 to ns2 np6 are collectively called. a) transition elements. b) representative element . c) lanthanides. d) inner transition elements Ans: b) representative element .
Q:8. Atomic numbers of V,Cr, Mn and Fe are 23,24,25and 26 respectively, which one of these may have the highest second ionisation potential a) Mn b) Fe c) v d) Cr Ans:- d) Cr E.c- of Cr – 3d5 ,4s1, as 3d- orbital is exactly half filled, removal of election is difficult.
Q: 9. Diagonal relationship is shown by a) all elements with their diagonally opposite elements. b) all elements of 3rd and 4th periods c) some elements of 2nd and 3rd periods. d) elements of d-block. Ans: c) some elements of 2nd and 3rd periods.
Q :10. Lithium is not stored in kerosene. It is kept wrapped in paraffin wax because a) it reacts with kerosene . b) Lithium floats over kerosene as its density is low c) Lithium is a soft metal. d) Lithium sinks in kerosene Ans : b) Lithium floats over kerosene as its density is low .
Q : 11. compounds of group 2 elements are less soluble in water than the corresponding alkali metal salts due to a) their high ionisation energy b) their high lattice energy c) less basic character d) low electronegativity values. Ans : b) their high lattice energy
Q :12. The Density of “Ca” is less than that of “Mg” because a) Nuclear charge of “Ca” is more than “Mg” b) Size of “Ca” is less than “Mg” c) Vacant 3d orbital of “Ca” d) completely filled 3d orbital’s . Ans: c) Vacant 3d orbital of “Ca” leads to the much increasing atomic volume hence density decreases
Q :13. Oxide of an element whose electronic configuration is 1s22s22p63s1 is a) Basic b) Acidic c) neutral d) Amphoteric Ans: a) Basic
Q : 14. carbon -60, contains a) 12 five membered and 20- six membered rings b) 12 five membered and 20- five membered rings c) Same no of five and six membered rings d) No five and six membered rings Ans: a) 12 five membered and 20- six memberd rings
Q :15. A radio station broadcasts on a frequency of 1368 KHz, what is the wavelength of electromagnetic radiation emitted by transmitter a) 219300m b) 219.3m c) 2.1933m d) 0.2193m Ans: b) 219.3m
C = 3 x 108 m/s λ = ? = = 3 x 108 1368 x 103 = 219.3 m
Q :16. Energy of a wave of wavelength 400A0 is a) 495 x 10-19J b) 4.95 x 10-18J c) 4.95 x 10-19J d) 495 x 10-18J Ans: c) 4.95 x 10-19J
λ= 400 A0 = 400 x 10-10 m E = h = h = 6.6 x 10-34 x 3 x 108 400 x 10-10 E = 4.95 x 10-19J
Q : 17. What will be de- Broglie wave length of an electron moving with a velocity of 1.2 x10-5m/s a) 6.068 x 10-9m b) 3.133 x 10-37m c) 6.626 x 10-9 m d) 6.018 x 10-7 m Ans: A) 6.068 x 10-9 m
λ= h/mc h = 6.62 x 10-34 js m = 9.1 x 10-31 Kg c = 1.2 x 105 m/s λ =
Q:18 . The Quantum numbers for the last electron of an element are given below n = 2, l = 0, m = 0, s = + 1/2 , The atom is a) Lithium b) Beryllium c) Hydrogen d) Boron Ans: a) Lithium
Q:19. Which of the following set of Quantum numbers is not possible n l m s a) 3 2 -1 0 b) 3 2 -1 + ½ c) 3 2 1 + ½ d) 3 2 1 - ½ Ans: a) n=3 l=2 m=-1 s=0
Q:20. Total number of neutrons in dipositive zinc ion with mass 70 is a) 34 b) 40 c) 36 d) 38 Ans b) 40 A = 70, z = 30 n = A-z = 40
Q:21. Orbital angular momentum of an electron in 2s orbital is A] + ½ h/2π B] h/2π C] √2 h/2π D] Zero Ans: d] Zero Orbital angular momentum = l h/2π For ‘S’ orbital l=0 0 x h/2π = 0
Q: 22. What is the wave number of radiation emitted when an electron jumps fro n =3 to n= 1 in a hydrogen atom (R=107/m) a) 1.11 x 108/m b) 8.9 x 108/m c) 8.9 x 106/md) 8.9 x 107/m Ans: c) 8.9 x 106/m
Q : 23. The ion that is isoelectronic with CO is a) b) c) d) CN‑ Ans : CN- No of electrons in co = 6+8 = 14 In CN- 6+7+1 = 14
Q: 24. It is impossible to measure simultaneously the position and momentum of a small particle with absolute accuracy, This is stated by a) Hund b) pauli c) Heisenberg d) Bohr Ans : c) Heisenberg
Q : 25. 4s orbital gets filled before 3d because a) 4s has more energy than 3d b) 4s has same energy energy as that 3d c) ( n + 1 ) Value for 4s is more than 3d d) ( n + 1 ) Value for 4s is less than 3d Ans: d) ( n + 1 ) Value for 4s is less than 3d