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標題:
polyatomic ion 的 o.n. 是否只有死記?
發問:
polyatomic ion 的 o.n. 是否只有死記? 因為有些可以跟佢帶幾多個charge 咁記埋.. but一定5係 all 都係咁.. 咁要記..是否只有死記?? (ps.. 本人學緊 redox reaction中的equation)
最佳解答:
當中有些的確要緊記，但記了這些後有很多都能算出來。 要記的是： 1. All elements have the o.n. zero when they exist in form of elements, such as H2, Al, Cl2 and so on. 2. Whenever these elements have the o.n. not equal to zero (usually are combined to form compounds), ... i) all group I elements have the o.n. +1 ii) fluorine has the o.n. -1 3. When these elements have the o.n not equal to zero (usually are combined to form compounds), ... i) hydrogen USUALLY has the o.n. +1 ii) oxygen USUALLY has the o.n. -2 exception : H2O2(hydrogen peroxide). o.n. of H is -1. o.n. of O is +1. 計算其他元素的 o.n. 的方法： 首先找出那個compound的 chemical formula, 如 rust is Fe3O4?xH2O、 dichromate ions are Cr2O7- ...... 第二，找出那個compound的 charge, 如 Fe3O4 is 0、 H2O is 0、 Cr2O7- is -1、 NH4+ is +1 …… 第三，寫出已肯定或唯一能肯定的 o.n. of the element, 如 Fe3O4：寫-2在O的上方、 Cr2O7-：寫-2在O的上方、 NH4+：寫+1在H的上方…… 第四，以此算式運算： (no. of the 1st element in the formula)*(the charge of the 1st element) + (no. of the last element in the formula)*(the charge of the last element) = charge of the compound Let y be the required o.n. 例： Fe3O4 : 3y + 4(-2) = 0 Cr2O7- : 2y + 7(-2) = -1 NH4+ : y + 4(+1) = +1 其實你不用太擔心，因為多點回答這類型的題目就很自然的 記得不少元素在某幾個情況下的 o.n.。那就不用常常運算。
其他解答:
polyatomic ion 的 o.n. 是否只有死記?
發問:
polyatomic ion 的 o.n. 是否只有死記? 因為有些可以跟佢帶幾多個charge 咁記埋.. but一定5係 all 都係咁.. 咁要記..是否只有死記?? (ps.. 本人學緊 redox reaction中的equation)
最佳解答:
當中有些的確要緊記，但記了這些後有很多都能算出來。 要記的是： 1. All elements have the o.n. zero when they exist in form of elements, such as H2, Al, Cl2 and so on. 2. Whenever these elements have the o.n. not equal to zero (usually are combined to form compounds), ... i) all group I elements have the o.n. +1 ii) fluorine has the o.n. -1 3. When these elements have the o.n not equal to zero (usually are combined to form compounds), ... i) hydrogen USUALLY has the o.n. +1 ii) oxygen USUALLY has the o.n. -2 exception : H2O2(hydrogen peroxide). o.n. of H is -1. o.n. of O is +1. 計算其他元素的 o.n. 的方法： 首先找出那個compound的 chemical formula, 如 rust is Fe3O4?xH2O、 dichromate ions are Cr2O7- ...... 第二，找出那個compound的 charge, 如 Fe3O4 is 0、 H2O is 0、 Cr2O7- is -1、 NH4+ is +1 …… 第三，寫出已肯定或唯一能肯定的 o.n. of the element, 如 Fe3O4：寫-2在O的上方、 Cr2O7-：寫-2在O的上方、 NH4+：寫+1在H的上方…… 第四，以此算式運算： (no. of the 1st element in the formula)*(the charge of the 1st element) + (no. of the last element in the formula)*(the charge of the last element) = charge of the compound Let y be the required o.n. 例： Fe3O4 : 3y + 4(-2) = 0 Cr2O7- : 2y + 7(-2) = -1 NH4+ : y + 4(+1) = +1 其實你不用太擔心，因為多點回答這類型的題目就很自然的 記得不少元素在某幾個情況下的 o.n.。那就不用常常運算。
其他解答:
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