4.
which element in group 17 (viia) is least likely to lose an electron?
a. chlorine
b. iodine
c. bromine
d. fluorine

The correct option is D)  fluorine

Explanation:

Fluorine has 9 electrons. Florine atom will have only two orbits in which the electrons will revolve around the nucleus. First orbit will have two electrons. The second orbit will have seven electrons. These orbits will be pulled by a greater force by the nucleus of an atom. Hence, they will be less likely to lose an electron.

Other atoms of the group 17 have more orbits hence, the space between the nucleus and the outermost orbit is greater than that of fluorine atoms. Hence, they will have more chances of losing an electron as compared to fluorine.

Fluorine

Explanation:

Remembering the equation for Coulombic attraction, we know that an electrostatic attraction between two charges is directly proportional to the product of the two charges and inversely proportional to the square of a distance between them. This is modeled by the equation:

The attraction force we are analyzing in this problem occurs between a nucleus of a halogen and its valence electrons. Essentially, according to the equation, the greater the distance between a nucleus and valence electrons, the lower the attraction force. This implies that a valence electron would be lost more easily from an atom which has the highest radius.

Going down the group, the number of electron shells increases and the radius of halogens increases. This means that fluorine has the smallest radius and attracts its valence electrons strongest.

under a microscope, cancer cells may look very different from normal cells. they often have different sizes and some may be larger than normal while others are smaller. cancer cells are often abnormally shaped and the control centre of the cell (the nucleus) may have an abnormal appearance.

i think? ?

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explanation: