answer bits 13 14 15 16 17 18 in the worksheet
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14. Elements having high electronegativity :
F 3.98
O. 3.44
CL. 3.16
Elements having low electronegativity :
Li. 0.98
Na. 0.98
K. 0.82
Rb. 0.82
Sr. 0.95
Cs. 0.79
Ba. 0.80
Fr. 0.7
Ra. 0.80
F 3.98
O. 3.44
CL. 3.16
Elements having low electronegativity :
Li. 0.98
Na. 0.98
K. 0.82
Rb. 0.82
Sr. 0.95
Cs. 0.79
Ba. 0.80
Fr. 0.7
Ra. 0.80
Hitler9322:
Elements having low electron affinity :
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14)
15) The number of protons and electrons is equal when the atom is in its natural state. But when it gets charged and forms an ion, it either loses or gains electrons but the number of protons remains unchanged. Hence, we depend on the proton to identify the element, even in its ionized form. Thus, the Atomic Number is the Number of Protons and NOT Electrons.
16)
a)2nd Period
b)The missing element is Oxygen (₈O), it should be placed between Nitrogen (₇N) and Florine (₉F).
c) Carbon (₆C) shows the property of Catenation.
d) Be,N,F
e) Florine (₈F) is an Halogen. It reacts aggressively with Lithium (₃Li) which is an Alkali.
17)
a) elements in group 1 are called Alkali
b) H, Li, Na, K, Rb, Cs, Fr
c)Caesium (Cs)
d)Alkali Metal Chlorides
Sorry, I could only answer this much. This is all I know. Sorry, i could not be of complete use to you
15) The number of protons and electrons is equal when the atom is in its natural state. But when it gets charged and forms an ion, it either loses or gains electrons but the number of protons remains unchanged. Hence, we depend on the proton to identify the element, even in its ionized form. Thus, the Atomic Number is the Number of Protons and NOT Electrons.
16)
a)2nd Period
b)The missing element is Oxygen (₈O), it should be placed between Nitrogen (₇N) and Florine (₉F).
c) Carbon (₆C) shows the property of Catenation.
d) Be,N,F
e) Florine (₈F) is an Halogen. It reacts aggressively with Lithium (₃Li) which is an Alkali.
17)
a) elements in group 1 are called Alkali
b) H, Li, Na, K, Rb, Cs, Fr
c)Caesium (Cs)
d)Alkali Metal Chlorides
Sorry, I could only answer this much. This is all I know. Sorry, i could not be of complete use to you
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