Yadda za a ƙayyade matsayi na samfur

All sinadaran abubuwa kunshi kwayoyin, wanda su ne mafi karami barbashi. Su ne tsaka tsaki electrically kuma an hada da kwayoyin halitta, wanda suna da nasaba da covalent shaidu. Rashin ikon samfurori na abubuwa sunadarai don samar da wasu adadin shaidu a cikin kwayoyin ana kiransa valence. Dangane da fitowar da kuma ci gaba da ka'idar sha'anin sinadarai, wannan ra'ayi ya shawo kan canje-canje mai mahimmanci kuma a halin yanzu ba shi da wani fassarar fassarar kimiyya, ana amfani dashi da yawa don dalilai na hanya kuma an gano shi da nauyin haɗari. Idan kwayoyin ita ce ƙananan ƙwayar wani abu, to, atomar ita ce ƙananan ƙananan nau'i na nau'i mai sinadarai. Har ila yau, ƙwayar ita ce matsala mai tsaka tsaki a fili, a tsakiya akwai tsakiya wanda ya kunshi protons, da kuma kewaye da shi akwai girgije na zaɓuɓɓuɓɓuka. Sabili da haka, cajin mai kyau na tsakiya ya daidaita ne ta hanyar cajin wutar lantarki mara kyau. Don fahimtar yadda za a ƙayyade matsayi na samfur da kuma abin da yake, kana buƙatar la'akari da tsarin kwayoyin na wasu abubuwa.

A cikin kwayoyin hydrogen chloride (HCl) tsakanin hydrogen da chlorine, akwai guda ɗaya na electrons. A hydrogen zarra yana daya proton da daya electron. Kwayar chlorine tana da nau'i 17 da 17 na lantarki, ɗaya daga cikin 7 na lantarki a cikin harsashi na asali na ma'aunin wutar lantarki ya haɗa tare da injin na iskar hydrogen, ta haka ne ke kafa nau'in lantarki: H: Cl. A cikin kwayar H2S, atomatik sulfur ya kafa nau'i biyu nau'i na electrons tare da dukkanin hydrogen: H: S: H. Yaya za a ƙayyade digiri na oxyidation na wani abu, misali, chlorine da hydrogen a cikin kwayoyin HCl da hydrogen da sulfur a cikin kwayar H2S? Valencia yana bada ra'ayi akan adadin shaidu ko nau'i-nau'i na lantarki wanda aka kafa a cikin kwayar, amma ba ya nuna ƙaurawarsu ba.

Ana tsammanin cewa an hada electrons na mahaifa a cikin kwayoyin zuwa nau'i-nau'i na lantarki, wanda aka canza zuwa mafi yawan ƙananan ƙarancin lantarki. Hydrogen da chlorine a HCl suna iya shiga cikin guda ɗaya, wanda aka canza zuwa atomatik chlorine, tun da yake shi ne mafi yawan matsala. Sabili da haka, a cikin hydrogen, mataki na oxyidation yana da 1, kuma ga chlorine, minus 1. Kowane ƙwayar hydrogen a cikin H2S kwayoyin ya zo cikin ɗaya bond tare da atomatik sulhu, wanda a biyun yana da biyu shaidu. Dukansu nau'i-nau'i guda biyu a cikin wannan yanayin an canza su zuwa ƙwayar sulfur. Sabili da haka, don hydrogen, mataki na daidaitaccen abu yana da 1, kuma ga sulfur atom - minus 2. Yaya za a ƙayyade digiri na oxyidation na wani abu, misali, HCl, H2S, O2, N2, Na, Ag, Si? Matsayin da samin abuwan abu mai sauƙi ko abu mai mahimmanci shine ko da yaushe 0, tun lokacin da abu ya ƙunshi kwayoyin ko mahaifa, kuma suna da tsaka tsaki a fili. Bisa ga wannan, an samo digiri na yin amfani da iskar shaka akan kashi a cikin kwayar.

Alal misali, tare da acid hypochlorous, wanda zai iya la'akari da yadda za a ƙayyade darajar samfurin oxyidation na atomin Cl. A cikin kwayar HClO, kamar kowane abu, hydrogen ko da yaushe (sai dai hydrides) yana da alamar samin oxyidation da 1, kuma an cire oxygen a kowane lokaci (2) (sai dai barium peroxide da oxygen fluoride). Tun da kwayoyin kanta ba ta da cajin, za mu iya rubuta daidaitattun: (+1) + x + (-2) = 0. Wannan daidaitattun dole ne a warware: x = 0 - (+1) - (-2) = (+1) , Saboda haka, ya juya cewa matsayin digirin samin na Cl a cikin kwayar hypochlorous acid ya daidaita da kuma 1. Wannan misali ya nuna yadda za a ƙayyade matsin lambar abu na musamman a kowace kwayoyin. Da farko dai, abubuwan da aka sani da waɗannan abubuwa waɗanda ba'a canzawa ba ne ko kuma digiri na rashin asali.

Saboda haka, hadawan abu da iskar shaka da ake kira da matukar daraja (m lura da zarra a cikin kwayoyin) amfani da rikodi lissafai halayen hade da hadawan abu da iskar shaka ko rage daban-daban abubuwa. Wannan darajar daidai yake da yawan nau'i-nau'i na zaɓuɓɓukan lantarki wanda aka juya gaba ɗaya ga nau'ikan kwayoyin halitta na kwayoyin halitta. Matsayin da aka yi amfani da iskar shaka an rubuta a sama da nuni na atomatik. Irin wannan rikodin ya nuna yadda za a ƙayyade matsin lambar samfur na wani abu a cikin kwayoyin kowane abu. Alal misali, a rikodin (Na + 1Cl-1) Dole ne a fahimci cewa gishiri kwayoyin sodium hadawan abu da iskar shaka digiri da 1, debe 1, da kuma chlorine kamar haka: (+1) + (-1) = 0. A kwayoyin na potassium permanganate (K + 1Mn + 7O-24) don potassium da 1, ga manganese da 7, domin oxygen minus 2, wato: (+1) + (+7) + 2 • (-2) = 0. A cikin kwayoyin nitric acid (H + 1N + 5O-23) da 1 ga hydrogen, domin nitrogen da 5 ga oxygen debe 2, da kuma dukan kwayoyin ne electrically tsaka tsaki: (+1) + (+5) + 3 • (-2) = 0. a cikin kwayoyin na nitrous acid (H + 1N + 3O-22) da 1 ga hydrogen, nitrogen domin da 3 oxygen debe 2, tun da lura da kwayoyin ne ya daidaita da sifili, sa'an nan: (1) + (+3) + 2 • (-2) = 0. A misali na nitrogen, ana iya ganin cewa wannan zarra zai iya samun jihohi daban-daban na abubuwa daban-daban daga 5 zuwa minus 3 (tare da wani lokaci na ɗaya ɗaya).